PHYS202 – SPRING 2009 - Delaware Physicsjdshaw/PHYS202/Lecture/Lecture_Notes... · total = 12+...

2202009

1

PHYS202 ndash SPRING 2009PHYS202 ndash SPRING 2009Lecture notes ‐ Electrostatics

2202009

2

What is Electricity

bull Static effects known since ancient times

bull Static charges can be made by rubbing certain materials together

D ib d b B j i F kli fl idbull Described by Benjamin Franklin as a fluidndash Excess of the fluid was described as positive

Deficit of the fluid was described as negativendash Deficit of the fluid was described as negative

bull Like charges repel unlike charges attract

2202009

3

A Quantitative Approach

bull Charles Augustin de Coulomb first described relationship of force and charges

bull Coulomb invented the torsion balance to measure th f b t hthe force between charges

bull Basic unit of charge in SI system is called the Coulomb

bull One of the basic irreducible measurements

ndash Like mass length and timetime

2202009

4

Fundamental Charge

bull Due to atomic structure

bull Positively charged heavy nuclei and negative charged electronsPositively charged heavy nuclei and negative charged electrons

bull Charged objects have excess or deficit of electrons

bull Protons and electrons have equal and oppositecharge e = 1602 x 10‐19 Cg

bull Rubbing some objects causes electrons tomove from one object to another

bull Negatively charged objects have an excessof electrons

bull Positively charged objects have a deficit ofelectrons

N t h i h i l d t tbull Net change in charge in a closed system mustbe zero

2202009

5

Fundamental Charge

Problem 186 ‐Water has a mass per molecule of 18 gmol and each molecule has 10 electrons How many electrons are ineach molecule has 10 electrons How many electrons are in one liter of water

23 310electrons 6022times10 molecules 1mol 1g 10 mltimes times1molecule 1mol 18g 1ml 1liter

N = times times

h h h f h l

26

gelectrons33 10

literN = times

What is the net charge of the electrons

( )( )26 19

7

33 10 electrons 1602 10 C electron

5 4 10 C

q Ne

q

minus= = times minus times

times54 10 Cq = minus times

2202009

6

Conservation of Charge

bull Net change in charge in any closed system is zero

bull Or the change of charge in any volume equals the net charge entering andOr the change of charge in any volume equals the net charge entering and leaving the system

( ) ( ) ( ) ( )Charge Original Charge Charge in Charge out= + minus

bull We denote the amount of charge by the letter Q or q

bull If Charge is created it must be equal and opposite

0 in out= + -q q q q

2202009

7


Two identical metal spheres of charge 50 C and ‐16 C are brought together and then separated What is the charge onbrought together and then separated What is the charge on each sphere

1 2 = 34Ctotalq q q= +

Since the charge is distributed equally each must have an equal charge

1 2totalq q q

equal charge

1 2 = 17C2totalqq q= =

2202009

8

Conductors and Insulators

bull Insulatorndash Charge stays where it is depositedCharge stays where it is depositedndash Wood mica Teflon rubber

C d tbull Conductorndash Charge moves ldquofreelyrdquo within the objectndash Metals exhibit high conductivity

bull Due to atomic structurendash Is the outermost electron(s) tightly bound( ) g y

2202009

9

Charging Objects

bull By frictionndash Rubbing different materialsndash Rubbing different materialsndash Example fur and ebonite

bull By contactB i i h d bj i i h hndash Bringing a charged object into contact with another (charged or uncharged) object

bull By inductionh b llndash Bringing a charged object near a conductor temporally

causes opposite charge to flow towards the charged objectGrounding that conductor ldquodrainsrdquo the excess chargendash Grounding that conductor drains the excess charge leaving the conductor oppositely charged

2202009

10

Force Between Charges

bull Coulomb found that the force between two charges isbetween two charges is proportional to the product of their charges and inversely proportion to the square of the distance between them

1 22

q qF prop

Where q1 and q2 represent the

212r

charges and r12 the distance between them

2202009

11

Coulombrsquos Law

bull By using the torsion balance Coulomb discovered the constant of proportionality k = 899 Nm2C2

1 22

q qF kr

=+ +

‐ ‐

bull If both q1 and q2 are both positive the force is positive

12r‐ +

bull If both q1 and q2 are both negative the force is positive

bull If both q1 is positive and q2 is negative the force is negative

bull Hence the sign of the force is positive for repulsion and negative for tt tiattraction

2202009

12

Coulombrsquos LawPrinciple of SuperpositionPrinciple of Superposition

bull Electric force obeys the law of linear superposition ndash the net force on a charge is the vector sum of the forces due to each charge

2î j

i iji j ij

q qk

rne

= sumF rr

bull Add the vectors

i j ijrne

1

3

The net force on charge4 is F14+F24+F34=Fnet

1

24

F14F24

F34

Red are positive chargesBlue are negative charges

2 24

2202009

13

Coulombrsquos Law Principle of SuperpositionPrinciple of Superposition

bull The forces add for as many charges as you have

2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36 The net force on charge 5 is the

vector sum of all the forces on

13

4

58

charge 5


24

7

2202009

15

Coulombrsquos Law1814 ndash Two tiny conducting spheres are identical and carry charges of ‐200 μC and1814 Two tiny conducting spheres are identical and carry charges of 200 μC and

+500 μC They are separated by a distance of 250 cm (a) What is the magnitude of the force that each sphere experiences and is the force attractive or repulsive

( )( )( )( )

9 2 2 6 61 2

22 2

899 10 N m C 200 10 C 500 10 Cq qF kminus minustimes sdot minus times times

= =

The net force is negative so that the force is attractive

( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times

(b) The spheres are brought into contact and then separated to a distance of 250 cm Determine the magnitude of the force that each sphere now experiences and state whether the force is attractive or repulsiveSi h h id i l h h h f b i d iSince the spheres are identical the charge on each after being separated is one‐half the net charge so q1 + q2 = +15μC

( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k

minus minustimes sdot times times= = = times

The force is positive so it is repulsive

( )2 2212 250 10 mr minustimes

2202009

16

Coulombrsquos Law1821 ndash An electrically neutral model airplane is flying in a horizontal circle on a 30 m

guideline which is nearly parallel to the ground The line breaks when the kinetic energy of the plane is 500 J Reconsider the same situation except that now there is a point charge of +q on the plane and a point charge of minusq at the other end of h id li I hi h li b k h h ki i f h l ithe guideline In this case the line breaks when the kinetic energy of the plane is 518 J Find the magnitude of the charges

From the definition of kinetic energy we see thatmv2 = 2(KE) so that Equation 5 3From the definition of kinetic energy we see that mv2 = 2(KE) so that Equation 53 for the centripetal force becomes

( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p

Centripetal forceforce

2202009

17

Coulombrsquos Law1821 ndash (cont)

( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243

Subtracting Equation (2) from Equation (1) eliminates Tmax and gives

force

( ) ( )2charged uncharged

2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=

Solving for |q| gives

( ) ( ) ( )( )charged uncharged ndash52 KE ndash KE 2 30 m 518 J ndash 500 J

35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18

The Electric Fieldbull Electric charges create and exert and experience the electric force

q q q2 2

ˆ î j ji ij i i ij

i j i jij ij

q q qk q k

r rne ne

rArr= =sum sumF r F rr r

bull We define the Electric Field as the force a test charge would feel if placed at that point

2ˆj

i j

qk q

rne

= rArr =sumE r F Er r r

bull The electric field exists around all charged objects

i jne

2202009

19

Electric Force Lines

bull Michael Faraday saw the force caused by one charge to becaused by one charge to be from lines emanating from the charges

bull Lines begin on Positive charges and end on Negative charges flowing smoothly through space never crossingspace never crossing

2202009

20


bull Faraday imagined these lines d h fand the force was greater

where the lines are denser weaker where they are spreadweaker where they are spread out

2202009

21

Electric Field Lines

bull Go out from positive charges and into ti hnegative charges

Positive Negative Neutral

2202009

22

The Inverse Square Law

bull Many laws based on the geometry of three‐di i ldimensional space

1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field

1828 ndash Four point charges have the same magnitude of 24 x 10‐12 C and are fixed to the corners of a square that is 40 cm on a side Three of the charges are positive d i ti D t i th it d f th t l t i fi ld th t i tand one is negative Determine the magnitude of the net electric field that exists

at the center of the squareB C++

ED

The drawing at the right shows each of the field contributions at the center of the square (see black d t) E h i di t d l di l f th

A D- +

EA

EC

EB

dot) Each is directed along a diagonal of the square Note that ED and EB point in opposite directions and therefore cancel since they have the same magnitude In contrast EA and EC point in the same direction toward corner A and therefore combine to give a net field thatcorner A and therefore combine to give a net field that is twice the magnitude of EA or EC In other words the net field at the center of the square is given by the following vector equation Σ = + + +A B C DE E E E E

2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED

The magnitude of the net field is A 22 2k q

E Er

Σ = =

A D- +

EA

EC

EB

In this result r is the distance from a corner to the center of the square which is one half of the diagonal distance d Using L for the length of a side of the square d t ki d t f th P th th

r

and taking advantage of the Pythagorean theorem we have With this substitution for r the magnitude of the net field becomes

2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =

times sdot timesΣ =

( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25


1834 ndash Review Conceptual Example 12 before attempting to work this problem The magnitude of each of the charges in Figure 18‐21 is 86 x 10‐12 C The lengths of th id f th t l 3 00 d 5 00 Fi d th it d f ththe sides of the rectangles are 300 cm and 500 cm Find the magnitude of the electric field at the center of the rectangle in Figures 18‐21a and b

1 21+ q - q

The figure at the right shows the configuration given in text Figure 1821a The electric field at the center of the rectangle is the resultant of the electric fields at the center due to each of the four charges As

344

+ q + q

of the electric fields at the center due to each of the four charges As discussed in Conceptual Example 11 the magnitudes of the electric field at the center due to each of the four charges are equal However the fields produced by the charges in corners 1 and 3 are in opposite directions Since they have the same magnitudes they combine to givedirections Since they have the same magnitudes they combine to give zero resultant The fields produced by the charges in corners 2 and 4 point in the same direction (toward corner 2) Thus EC = EC2 + EC4 where EC is the magnitude of the electric field at the center of the rectangle and EC2 and EC4 are the magnitudes of the electric field at the C2 C4center due to the charges in corners 2 and 4 respectively Since both EC2 and EC4 have the same magnitude we have EC = 2 EC2

2202009

26

Electric Field Lines1834 ndash (cont)

The distance r from any of the charges to the center of the rectangle can be found using the Pythagorean theorem

1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =

2Therefore 292 cm 292 10 m2dr minus= = = times

9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus

The figure at the right shows the configuration given in text Fi 18 21b All f h t ib t t

9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus

times sdot minus times= = = = minus times

times

- q - q

Figure 1821b All four charges contribute a non‐zero component to the electric field at the center of the rectangle As discussed in Conceptual Example 11 the contribution from the charges in corners 2 and 4 point toward corner 2 and the contribution from the charges in corners 1 and 3 point toward corner 1

1 2

34

1

4

C

E 13 E 24

the charges in corners 1 and 3 point toward corner 1 344

+ q + q

2202009

27


Notice also the magnitudes of E24 and E13 are equal and from the first part of this problem we know that

E24 = E13 = 181 times 102 NC

1 21- q - q

E 13 E 243The electric field at the center of the rectangle is the vector sum of E24 and E13 The x components of E24 and E13 are equal in magnitude and opposite in direction hence

(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

From Figure 2 we have that

500 cm 500 cmsin 0858583 cmd

θ = = =

( ) ( )( )2 2C 132 sin 2 181 10 NC 0858 311 10 NCE E θ= = times = times

2202009

28

The Electric Field


bull This is shown byelectric fieldelectric fieldlines

bull Constructedid i ll hidentically to theforce lines

2202009

29

Drawing Electric Field Lines

bull Number of lines at a charge is proportional to the chargethe charge

2202009

30


bull The charge with ‐3qmust have three times as many lines as the charge with +q and 15 times the lines for +‐ 2qthe charge with q and 15 times the lines for 2q

2202009

31

Parallel Plates of Charge

bull Adding fields from each point on surface causes field lines parallel to the plates to cancel

PositivelyCharged

NegativelyCharged

bull Away from the edges the field is parallel and perpendicular to the plate(s)the plate(s)

2202009

32

Electric Field amp Conductors

bull In conductors excess electrons will move when an electric field is present

bull For ldquosteady staterdquo (ie no moving charges) the net electric field must be zeror r r r

bull Therefore the charges move to the surface(s) of the conductor until they cancel the electric field inside the conductor

bull Field lines can start and end on a conductor at right angles to the surface of the

0q rArr= = =F E F Er r r r

Field lines can start and end on a conductor at right angles to the surface of the conductor

2202009

33


1844 ndash Two particles are in a uniform electric field whose value is + 2500 NC The mass and charge of particle 1 are m1 = 14 x 10‐5 kg and q1 = ‐70 μC while the

di l f ti l 2 2 6 10 5 k d 18 C I iti llcorresponding values for particle 2 are m2 = 26 x 10‐5 kg and q2 = +18 μC Initially the particles are at rest The particles are both located on the same electric field line but are separated from each other by a distance d When released they accelerate but always remain at this same distance from each other Find d

Ex

d

+x

q1 = minus70 μC

m1 = 14 times 10minus5 kg

q2 = +18 μC


The drawing shows the two particles in the electric field Ex They are separated by a distance d If the particles are to stay at the same distance from each other after being released they must have the same acceleration so ax1 = ax2 According to Newtonrsquos second law the acceleration ax of each particle is equal to the net force x p qΣFx acting on it divided by its mass m or x xa F m= Σ

2202009

34


1844 ndash (cont)

N t th t ti l 1 t b t th l ft f ti l 2 If ti l 1 t th i ht

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


Note that particle 1 must be to the left of particle 2 If particle 1 were to the right of particle 2 the particles would accelerate in opposite directions and the distance between them would not remain constant

1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =

Note the signs indicated the direction of the vector pointing from charge 2 to charge 1 for the force on charge 1

2202009

35


1844 ndash (cont)

Th l ti f h h t b th if th di t b t th

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


The acceleration of each charge must be the same if the distance between the charges remains unchanged

1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36

Electric Fluxbull Flux (from Latin ndash meaning to flow)

bull Electric flux measures amount of electric field passing through a surface

Er

Er

E E

S t l l t fi d th l t

φ

Flux through surface is Area of surface times the field flowing through that surface

So to calculate find the normal to the surface

Then find the angle between the field and that normal vector φ

E EAΦ =field and that normal vector φ

( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er

This can be expressed as the vector dot product

Φ E Arr E


EΦ = sdotE A

cosE A φsdot =E Arr



cosE Aor

φsdot =E A

field and that normal vector φ

x x y y z zE A E A E Asdot = + +E Arr

2202009

38

Electric Flux

1848 ndash A rectangular surface (016 m x 038m) is oriented in a uniform electric field of 580 NC What is the maximum possible electric flux through the surface

The maximum possible flux occurs when the electric field is parallel to the normal of the rectangular surface (that is when the angle between the direction of the field and the direction of the normal is zero) Then

( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39

Gaussrsquo Lawbull Electric Flux through a closed surface is proportional to the charge enclosed by that surface

encE

qΦ =

bull The fundamental constant ε0 is called the permittivity of free space It is related to Coulombrsquos law by ε0 = 885 x 10‐12 C2(Nm2)

bull Gauss Law is most useful when the electric field on the entire surface is constant

0ε

0

14

kπε

=

bull Take a point charge Construct a sphere centered on the point charge (all points on the surface equidistant from the charge) ndash the electric field is constant

bull Since the field is radial it is perpendicular to the concentric spherical surface everywhere sobull Since the field is radial it is perpendicular to the concentric spherical surface everywhere so φ = 0 cos φ = 1

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

bull The direction of the field is radially outward from the charge (or inward towards the charge if q is negative) ( )2

20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40

Gaussrsquo Law for a plate of chargebull In this case we take our closed surface to be a box oriented such that two faces are

parallel to the plate of charge and the others are perpendicular to the plate

Si th l t i fi ld f l t f h i di l t th f f thbull Since the electric field from a plate of charge is perpendicular to the surface of the plate the sides that are perpendicular to the plate have no flux

bull The charge enclosed by the box is the surface

Er

the box is the surface charge density multiplied by the area the box passes through the plate Aσthe plate

bull The area of the box with flux is 2A

Aσ

0 0 0E

q AEA Eσ σε ε ε

rArr rArrΦ = = =

2202009

41

Gaussrsquo Law for a line of charge

1854 ndash A long thin straight wire of length L has a positive charge Q distributed uniformly along it Use Gaussrsquo law to find the electric field created by this wire at a di l di tradial distance r

As the hint suggests we will use a cylinder as our Gaussian surface The electric field must point radially outward from the wire Because of this symmetry the field must be the same strength pointing radially outward on the surface of the cylinder regardless of the point on the cylinder thus we are justified in usingcylinder regardless of the point on the cylinder thus we are justified in using Gaussrsquo Law

Long straight wire Gaussian cylinderGausssiancylinderE

++++++ ++

12

3

r

L

End view

2202009

42


1854 ndash (cont) Let us define the flux through the surface Note that the field is parallel to the

d th th l b t th l f th f ( hi h i ll l tendcaps thus the angle between the normal from the surface (which is parallel to the wire) is perpendicular to the field thus φ = π2 so cos(φ)=0 for the ends For the side if the cylinder the flux is

( )2E sideEA E rLπΦ = =Long straight wire Gaussian cylinder

Thus from Gaussrsquo Law

( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=

λIt is customary to refer to charge density λ = QL

End view02

Er

λπε

=

2202009

43

Work

bull Let us take an object with charge q in a constant electric field E

bull If we release this charge the electric field causes a force to move the charge

bull Let us define the field as so there is only one dimension to consider Let us also set the initial velocity and position to be both zero

qqm

= rArr =F E a E

Ê=E xus also set the initial velocity and position to be both zero

bull Let the object move through a distance d The equations of motion are

21x at at= =υ

bull So after a time t the object moves 2

x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm

= rArr =υ υ2 2d ad

a a= rArr =υ

υ

2202009

44

Conservation of Energy

bull The left hand side is the kinetic energy the particle has after moving a distance d

212

m qEd=υ

bull That means the right hand side must represent some form of potential energy the object had prior to being released

bull Types of Energy already covered

ndash Translational kinetic energy

ndash Rotational kinetic energy

ndash Gravitational potential energy

ndash Elastic (spring) potential energy

2 2212

1 )12

(2totalE kx Em h PI g Emω= + + + +υ

bull Electric Potential Energy

ndash Cutnell amp Johnson use (EPE) however it is often more convenient to use U

2 22

2202009

45

Electric Potential Energy

bull Like gravitational potential energy EPE is defined with respect to the energy at some location

bull Importance lies in the difference between electric potential energy between two points

2202009

46

Example Problem

194 ndash A particle has a charge of +15 μC and moves from point A to point B a distance of 020 m The particle experiences a constant electric force and its motion is l th li f ti f th f Th diff b t th ti l rsquo l t ialong the line of action of the force The difference between the particlersquos electric potential energy at A and B is UA ndash UB = +90 x 10‐4 J Find the magnitude and direction of the electric force that acts on the particle and the magnitude and direction of the electric field that the particle experiences The work done must be equal to the change in potential energyThe work done must be equal to the change in potential energy

(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus

minus times= = = times

rArr

(from A to B)

But we know that for a particle in an electric field F = qE

( )45 10 N

020mF

xtimes

Δ

A B A BU U U UqE Ex q xminus minus

= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential

bull The electric potential energy of an object depends on the charge of an object

( )EPEbull Once again we can find a measurement independent of the objects charge

bull The SI unit for V is measured in Volts ( = JoulesCoulomb)

( )EPEV

q=

bull A common unit of energy is the electron Volt

bull Specifically How much potential energy does an electron get if moved through a potential of 1 Volt

191eV 1602 10 Jminus= times

2202009

48

Electric Potential

bull Like the Electric field the electric potential is independent of any test charge placed to measure a force

bull Unlike the electric field the potential is a scaler not a vector

bull By using calculus methods it can be shown that the electric potential due to aBy using calculus methods it can be shown that the electric potential due to a point charge is

N h h l f h l i i l d d h i f h h

kqVr

=

bull Note that the value of the electric potential depends on the sign of the charge

2202009

49

Electric Potentialfor a point chargefor a point charge

bull Example What was the potential difference due to a charge qA = 100 nC at the distances r0 = 100 cm and r1 = 200 cm qA 100 nC at the distances r0 100 cm and r1 200 cm

0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞

Δ = minus⎜ ⎟⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V

minus ⎛ ⎞Δ = times sdot minus⎜ ⎟⎝ ⎠

Δ = 450VVΔ

2202009

50


bull Example A charge qA is held fixed A second charge qB is placed a distance r0 away and is released What is the velocityplaced a distance r0 away and is released What is the velocity of qB when it is a distance r1 away

bull We use the conservation of energybull We use the conservation of energy

2 2 20 0 1 1

1 1 12 2 2

Energy m U m U m U= + rArr + = +υ υ υ

( )21 0 1 0 1

12 Bm U U q V V= minus = minusυ

2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ

bull Find the electric potential at r0 and r1

0 1A Akq kqV V= =

m

0 10 1

V Vr r

122 1 1A A A Bkq kq kq qq

m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52


bull For a single charge we set the potential at infinity to zero

kqV =

bull So what is the potential energy contained in two charges

bull Bring in the charges from infinity

bull The first charge is free as it does not experience any forces

Vr

=


bull The second moves through the field created by the first

bull So the energy in the system must be

2

1 1 1

12 12

rU q V V V Vkq kq kqVr r

infinΔ = Δ Δ = minus

Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53

Potential Energyof a collection of chargesof a collection of charges

bull Letrsquos add a third charge to this group

2 1 3 1 3 2

1 21 2

13 23

U q V q V q Vkq kqV Vr r

Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

bull Each successive charge experiences the field (ie potential) created by the previous charges Be careful not to double countcount

2202009

54

Example Problem

1913 ndash An electron and a proton are initially very far apart (effectively an infinite distance apart) They are then brought together to form a hydrogen atom in which th l t bit th t t di t f 5 29 10 11 Wh t i ththe electron orbits the proton at an average distance of 529 x 10‐11 m What is the change in the electric potential energyThe first charge comes in for free the second charge experiences the field of the first so

2 rU q V V V VinfinΔ = Δ Δ = minus

( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin

times times( ) ( )( )19

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus

times sdot timesΔ = minus times

timesΔ = minus times

2202009

55

Electric Field and Potential

bull Let a charge be moved by a constant electric field from points A to B (let Δd be that distance)

W F d W qE d= Δ rArr = Δ

bull Now we know that the work done must be equal to the negative change in electric potential energy


p gy

bull But potential energy is simply

ABU qE dminusΔ = Δ

bull Thus

ABq V qE dminus Δ = Δ

VE Δ=E

d= minus

Δ

2202009

56


bull This implies that the field lines are perpendicular to lines (surfaces) of equal

VEd

Δ= minus

Δ

potential

bull This equation only gives the component of the electric field along the displacement Δd p

bull To truly find the electric vector field you must find the components in each direction

V V VΔ Δ Δx y z

V V VE E Ex y z

Δ Δ Δ= minus = minus = minus

Δ Δ Δ

2202009

57


bull If the potentials at the following points (x y) areV(11) = 30 V V(091)=15V and V(109)=40 V What is theV(11) 30 V V(091) 15V and V(109) 40 V What is the (approximate) electric field at (11)

x y zV V VE E Ex y z


Δ Δ Δ

30V 15V 150 V m10m 09mxE minus

= minus = minusminus

30V 40V 100 V m10m 09myE minus

= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example

Problem 1934 ndash Determine the electric field between each of theelectric field between each of the points

VEx

Δ= minus

ΔxΔ

50V 50V 0 V m020m 00mABE minus

= minus =minus

30V 50V 100 V m040m 020mBCE minus

= minus =minus

10V 30V 50 V m080m 040mCDE minus

= minus =minus

2202009

59

Equipotential Surfaces

bull Electric field lines are perpendicular to equipotential surfaces

bull Electric field lines are perpendicular to surfaces of conductors

bull Thus surfaces of conductors are equipotential surfaces

bull The interior of conductors have no electric field and are therefore equipotential

bull The Electric potential of a conductor is constant throughout the conductor

2202009

60

Examplef1968 ndash An electron is released at the negative plate of a

parallel plate capacitor and accelerates to the positive plate (see the drawing) (a) As the electron gains kinetic energy does its electric potential energy increase or decrease Whydecrease Why The electric potential energy decreases The electric force F is a conservative force so the total energy (kinetic energy plus electric potential energy) remains constant as the electron moves across the capacitor Thus as the electron accelerates and its kinetic energy increases its electric potential energy decreases

(b) The difference in the electronrsquos electric potential energy between the positive and negative plates is Upos ndash Uneg How is this difference related to the charge on the electron (minuse) and to the difference in the electric potential between the platesThe change in the electronrsquos electric potential energy is equal to the charge on theThe change in the electron s electric potential energy is equal to the charge on the electron (ndashe) times the potential difference between the plates

(c) How is the potential difference related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plateand the displacement of the positive plate relative to the negative plate

2202009

61

Example( ) ( ) ff1968 ndash (cont) (c) How is the potential difference related to

the electric field within the capacitor and the displacement of the positive plate relative to the negative plate

The electric field E is related to the potential difference between the plates and the displacement Δs by

( )pos negV VE

s

minus= minus

Δ

Problem The plates of a parallel plate capacitor are separated by a distance of 12 cm

Note that (Vpos ndash Vneg) and Δs are positive numbers so the electric field is a negative number denoting that it points to the left in the drawing

and the electric field within the capacitor has a magnitude of 21 x 106 Vm An electron starts from rest at the negative plate and accelerates to the positive plate What is the kinetic energy of the electron just as the electron reaches the positive plateUse conservation of energy

V VKE U KE U KE KE

Total energy at Total energy atpositive plate negative plate

pos pos neg n pos negeg pos negeV eVKE U KE U KE KErArr + = ++ = +1442443 1442443

2202009

62

Example1968 ndash (cont)

But the electron starts from rest on the positive plate so h ki i h i i l i

pospos neg negKE KEeV eV+ = +

the kinetic energy at the positive plate is zero

pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=

But so( )pos negV VE

minus

( )pos n gneg ee V VKE minus=

But so( )E

s= minus

Δ

( ) ( )( )( )19 61602 10 C 216 10 V m 0012mnegKE se E minusminus Δ == minus times minus times

1540 10 JnegKE minustimes=

2202009

63

Lines of Equal Potential

bull Lines of equal elevation on amap show equal potentialmap show equal potentialenergy (mgh)

2202009

64

Example

1929 ndash Two equipotential surfaces surround a +15 x 10‐8 C point charge How far is the 190‐V surface from the 750‐V surfaceTh l t i t ti l V t di t f i t h i V k ThThe electric potential V at a distance r from a point charge q is V = kqr The potential is the same at all points on a spherical surface whose distance from the charge is r =kqV The radial distance r75 from the charge to the 750‐V equipotential surface is r75 = kqV75 and the distance to the 190‐V equipotentialsurface is r = kqV The distance between these two surfaces issurface is r190 = kqV190 The distance between these two surfaces is

75 19075 190 75 190

1 1kq kqr r kqV V V V

⎛ ⎞minus = minus = minus⎜ ⎟⎜ ⎟

⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC

r r minus⎛ ⎞sdot ⎛ ⎞minus = times + times minus =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

2202009

65

Capacitors

bull A capacitor consists of two conductors at different potentials being used to store charge

bull Commonly these take the shape of parallel plates concentric spheres or concentric cylinders but any conductors will do

bull q = CV where C is the capacitance measured in SI units of Farads (Coulomb Volt or C2 J ) V is the voltage difference between the conductors and q is the charge on each conductor (one is positive the other negative)

bull Capacitance is depends only upon the geometry of the conductors

2202009

66

Parallel Plate Capacitor

bull Field in a parallel plate capacitor is0

qEAε

=

bull But the a constant electric field can be defined in terms of electric potential

V

0

bull Substituting into the equation for capacitance

VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors

bull Two concentric spherical shells

4πε041 1

C

a b

πε=⎛ ⎞minus⎜ ⎟⎝ ⎠

bull Cylindrical capacitor has a capacitance perunit length 2C πε02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68

Energy stored in a capacitor

bull A battery connected to two plates of a capacitor does work charging the capacitorcharging the capacitor

bull Each time a small amount of charge is added to the capacitor the charge increases so the voltage between the platesthe charge increases so the voltage between the plates increases

V

qc V =qc

qΔqq

2202009

69


bull Potential Energy is U = qV

bull We see that adding up each small increase in charge Δq x V is just the area of the triangle 1

2U qV q CV= =

Such that2

212 2qU U CVV

= =

VSo the energy density (i e energy per unit

qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

So the energy density (ieenergy per unit volume) must be

qΔqq

( )02U E Adκε=

Ad is the area of the capacitor2

012

Uu EAd

κε= =

2202009

70

Example

bull How much energy is stored in a parallel plate capacitor made of 100 cm diameter circles kept 20 mm apart when aof 100 cm diameter circles kept 20 mm apart when a potential difference of 120 V is applied

bull Find the capacitancebull Find the capacitance2

0 0A rCd dε ε π

= =

bull The energy stored is

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd

πε πminussdot sdot

= = =( )

9209 10 JU minus= times

2202009

71

Electric Polarization in Matter

bull Electric charge distribution in some molecules is not uniformis not uniform

bull Example Water ndash both HExample Water both Hatoms are on one side

bull In the presence of anElectric Field water willalign opposite to the fieldalign opposite to the field

2202009

72


bull In the absence of a field l l ll bMolecules will be

oriented randomly

bull In an electric field they line up opposite to theline up opposite to the field

2202009

73


bull The resulting electric field is the sum of the external field and the fields generated by the molecules

bull The electric field strength through the matter is reduced

bull The reduction of the field depends on the energy in the molecules the strength of the field each one makes and the density of thefield each one makes and the density of the molecules

bull This can be measured and the material is said to have a dielectric constant κto have a dielectric constant ‐ κ

2202009

74

Dielectrics

bull This dielectric constant changes the value of the permittivity in space (ε)

q qbull Coulombs law in a dielectric is

bull Where ε is related to ε by a constant κ

1 22

12

ˆ4q q

rπε=F r

ε κε=bull Where ε is related to ε0 by a constant κ

bull The dielectric constant for a vacuum is κ = 1

0ε κε=

bull All other materials have a dielectric constant value greater than 1

bull Water at about 20˚C has κ = 804

bull Air at about 20˚C has κ 1 00054bull Air at about 20 C has κ = 100054

2202009

75

Dielectrics

0ε κε=

bull How does this affect the capacitance of parallel plates

AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76

Dielectric Breakdown

bull If the field in the dielectric becomes too t (i lt i t t) th t i lstrong (ie voltage is too great) the material begins to conduct current ndash this is called dielectric breakdowndielectric breakdown

bull When this happens the charge flows throughWhen this happens the charge flows through the material to the other plate and neutralizes the capacitor

2202009

77

Example1957 ndash Two hollow metal spheres are concentric with each other The inner sphere

has a radius of 01500 m and a potential of 850 V The radius of the outer sphere is 01520 m and its potential is 820 V If the region between the spheres is filled with Teflon find the electric energy contained in this spaceThe electric energy stored in the region between the metal spheres is

2 20 0

1 1 ( )2 2

u E E VolumU eκε κεrArr ==

The volume between the spheres is the difference in the volumes of the spheres

( )3 3 3 32 2 2 2

4 4 4( )3 3 3

Volume r r r rπ π π= minus = minus

Since E = ΔVΔs and Δs = r2 ndashr1 so E = ΔV(r2 ndashr1)

( )2

3 3 81 4 1 2 10 JV r rU κε π minus⎛ ⎞Δ ⎡ ⎤minus = times⎜ ⎟ ⎢ ⎥= ( )0 2 22 1

12 10 J2 3

r rr

Ur

κε π minus = times⎜ ⎟ ⎢ ⎥minus ⎣ ⎦⎝ ⎠=

2202009

2

What is Electricity

bull Static effects known since ancient times

bull Static charges can be made by rubbing certain materials together

D ib d b B j i F kli fl idbull Described by Benjamin Franklin as a fluidndash Excess of the fluid was described as positive

Deficit of the fluid was described as negativendash Deficit of the fluid was described as negative

bull Like charges repel unlike charges attract

2202009

3







2202009

4

Fundamental Charge









2202009

5

Fundamental Charge



N = times times

h h h f h l

26

gelectrons33 10

literN = times


( )( )26 19

7


5 4 10 C

q Ne

q



2202009

6








2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

3







2202009

4

Fundamental Charge









2202009

5

Fundamental Charge



N = times times

h h h f h l

26

gelectrons33 10

literN = times


( )( )26 19

7


5 4 10 C

q Ne

q



2202009

6








2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

4

Fundamental Charge









2202009

5

Fundamental Charge



N = times times

h h h f h l

26

gelectrons33 10

literN = times


( )( )26 19

7


5 4 10 C

q Ne

q



2202009

6








2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

5

Fundamental Charge



N = times times

h h h f h l

26

gelectrons33 10

literN = times


( )( )26 19

7


5 4 10 C

q Ne

q



2202009

6








2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

6








2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

7





1 2totalq q q

equal charge


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

8





2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

9

Charging Objects





2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

10



1 22

q qF prop


212r


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

11

Coulombrsquos Law


1 22

q qF kr

=+ +

‐ ‐


12r‐ +




2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

12



2î j

i iji j ij

q qk

rne

= sumF rr


i j ijrne

1

3


1

24

F14F24

F34


2 24

2202009

13



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2î j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

13



2ˆi j

i iji j ij

q qk

rne

= sumF rr

i j ijrne

36


13

4

58

24

7

2202009

14



2ˆi j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

14



2ˆi j

i iji j ij

q qk

rne

= sumF rr

i j ijrne



13

4

58

charge 5


24

7

2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

15



( )( )( )( )

9 2 2 6 61 2

22 2


= =


( )22 212

4

250 10 m

144 10 N

r

F

minustimes

= minus times


( )( )( )9 2 2 6 631 2

2 2

899 10 N m C 150 10 C 150 10 C324 10 N

q qF k




2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

16




( )2

c2 KEmF

r r= =υ

Sor r

( )( )

( )( )

2charged uncharged

max max2Centripetal

C t i t l

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243 p


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

17


( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forcef

2 KE 2 KE1 2

k qT T

r rr+ = =

1424314243


force


2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



35 10 Cr

q⎡ ⎤⎣ ⎦= = = times9 2 2 35 10 C

899 10 N m Cq

ktimes

times sdot

2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

18


q q q2 2


i j i jij ij

q q qk q k

r rne ne



2ˆj

i j

qk q

rne



i jne

2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

19




2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

20




2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

21




2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

22



1 22g

Gm mFr

= minus

2 24 4sound light

rP LI Ir rπ π

= =

1 22e

q qF kr

=

2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

23

The Electric Field



ED


A D- +

EA

EC

EB


2

= + + minus

= +

=

A B C B

A C

A

E E E EE EE

2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

24

The Electric Field

1828 ndash (cont)

B C++

ED


E Er

Σ = =

A D- +

EA

EC

EB


r


2r L=

4k k

( )( )( )

2 2

9 2 2 12

42

2

4 899 10 N m C 24 10 C

k q k qE

LL

Eminus

Σ = =


( )20040 m

54 NC

E

E

Σ =

Σ =

2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

25



1 21+ q - q


344

+ q + q


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

26



1 21

d500 cm2 2(3 00 ) +(5 00 ) 5 83d

344

300 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 122 2(8 99 10 N C )( 8 60 10 C)k minus


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er minus


times

- q - q


1 2

34

1

4

C

E 13 E 24


+ q + q

2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

27



E24 = E13 = 181 times 102 NC

1 21- q - q


(E13)x ndash (E24)x = 0

344

+ q + q

C

13 x 24 x

Therefore

F Fi 2 h th t

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =


500 cm 500 cmsin 0858583 cmd

θ = = =


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

28

The Electric Field




2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

29



2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

30



2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

31



PositivelyCharged

NegativelyCharged


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

32








2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

33




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

34


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2 1 1 2x x

q qF q E k

dΣ = minus 1 2

2 2 2x xq q

F q E kd

Σ = + 1 1 2

1 21 2 1

11 1

x x

xxx

dq q

q E kF dam m

minusΣ= =

2 2 2

1 22 2 2

22 2

x x

xxx

dq q

q E kF dam m

+Σ= =


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

35


1844 ndash (cont)


Ex

d

+x

q1 = minus70 μC


q2 = +18 μC



1 2x xa a

q q q q

=

Solving for d

1 2 1 21 22 2

1 2

x xq q q q

q E k q E kd d

m m

minus +=

Solving for d

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

36



Er

Er

E E


φ





( )cosE E AφΦ =

2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

37

Electric Flux

( )cosE E AφΦ =

Er


Φ E Arr E


EΦ = sdotE A




cosE Aor

φsdot =E A



2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

38

Electric Flux



( ) ( )( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

39


encE

qΦ =



0ε

0

14

kπε

=



0 0

cosenc encE

q qEA φε ε

Φ =rArr=


20 0

44

encq qE r Er

πε πε

= rArr =

2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

40





Er



Aσ

0 0 0E


rArr rArrΦ = = =

2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

41





++++++ ++

12

3

r

L

End view

2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

42






( )E side

++++++ ++

12

3

L

r

( )2encE

q QE rLπΦ = =rArr

GausssiancylinderE

( )0 0

02QE

rL

ε ε

πε=


End view02

Er

λπε

=

2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

43

Work




qqm

= rArr =F E a E



21x at at= =υ


x at atυ

x d=

1E

21 2 2

d at t d a= rArr =

2

t a

d

= υ

υ 2122

qE d m qEdm


a a= rArr =υ

υ

2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

44



212

m qEd=υ







2 2212

1 )12




2 22

2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

45




2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

46

Example Problem


(from A to B)4

390 10 J 45 10 N

A B A B

A B

Work U U F x U U

U UFminus

minus

= minus Δ = minus


rArr

(from A to B)


( )45 10 N

020mF

xtimes

Δ


= =Δ Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

x q x

Eminus

minus

Δ Δ

= = timestimes

times

2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

47

Electric Potential




( )EPEV

q=




2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

48

Electric Potential





kqVr

=


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

49



0 10 1

A Akq kqV V Vr r

Δ = minus = minus

0 1

1 1AV kq

r r⎛ ⎞


( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02m

450V

V

V


Δ = 450VVΔ

2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

50




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


2q ( )1 0 12q V Vm

= minusυ

2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

51


( )1 0 12q V Vm

= minusυ


0 1A Akq kqV V= =

m

0 10 1

V Vr r


m r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

bull Or

0 1 0 1m r r m r r⎝ ⎠ ⎝ ⎠

1 2 q V m= Δυ

2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

52



kqV =




Vr

=




2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

12 12r rinfin

2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

53



2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =13 23r r

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

54

Example Problem



( )( )

2

1 1 1

12 12

9 2 2 198 99 10 N m C 1 602 10 C

rqkq kq kqVr r

infin

minus

Δ = minus =infin


11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J

U

U

minusminus

minus



2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

55






p gy



bull Thus


VE Δ=E

d= minus

Δ

2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

56



VEd

Δ= minus

Δ

potential



V V VΔ Δ Δx y z

V V VE E Ex y z


Δ Δ Δ

2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

57





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟⎝ ⎠

o

x

E

(11)

y

150Vminus⎝ ⎠

2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

58

Example


VEx

Δ= minus

ΔxΔ


= minus =minus


= minus =minus


= minus =minus

2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

59







2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

60





2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

61




( )pos negV VE

s

minus= minus

Δ




V VKE U KE U KE KE



2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

62





pos neg

pos

neg

neg neg

KE

KE

eV eV

eV eV

= +

minus=


minus


But so( )E

s= minus

Δ



2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

63



2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

64

Example


75 19075 190 75 190



⎝ ⎠

( )2

9 875 190 2

N m 1 1899 10 150 10 C 11 m750 V 190 VC


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

65

Capacitors





2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

66



qEAε

=


V

0


VE V Edd

= minus rArr =

0

q CV CEd

AqC d C ε

= =

⎛ ⎞rArr⎜ ⎟ 0

0

qq C d CA dε

= rArr =⎜ ⎟⎝ ⎠

2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

67

Other Capacitors


4πε041 1

C

a b



ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

68




V

qc V =qc

qΔqq

2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

69




2U qV q CV= =

Such that2

212 2qU U CVV

= =


qc V =qc ( )

( )

20

2

121

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=


qΔqq

( )02U E Adκε=


012

Uu EAd

κε= =

2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

70

Example



0 0A rCd dε ε π

= =


( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =( )


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

71





2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

72



oriented randomly


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

73






2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

74

Dielectrics




1 22

12

ˆ4q q

rπε=F r



0ε κε=




2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

75

Dielectrics

0ε κε=


AA κεε 0

0

d

d

AACd d

C C

κεε

κ

= =

=

2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

76




2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


2202009

77



2 20 0

1 1 ( )2 2



( )3 3 3 32 2 2 2

4 4 4( )3 3 3



( )2


12 10 J2 3

r rr

Ur


PHYS202 – SPRING 2009 - Delaware Physicsjdshaw/PHYS202/Lecture/Lecture_Notes... · total = 12+...

Documents

Transcript of PHYS202 – SPRING 2009 - Delaware Physicsjdshaw/PHYS202/Lecture/Lecture_Notes... · total = 12+...