Electrics Review

8/3/2019 Electrics Review

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Electrics Review

Electrostatics Review

Electric Potential Review

Electric Flux & Gaussian Theorem Review

Electric Capacitance Review

Akshay

Satagopan

Akshay

Satagopan

2011.09.25

16:30:53 +05'30'



Electrostatics

Review



Frictional Electricity:

Frictional electricity is the electricity produced by rubbing two sui

and transfer of electrons from one body to other.

.

+ +

+ +

+ +

+ +

+ +

+ +

- -

- -

- -

- -

+ + +

+ + + +

+ + +

Glass

Silk

Electrons in glass are loosely bound in it than the electrons in silglass and silk are rubbed together, the comparatively loosely boufrom glass get transferred to silk.As a result, glass becomes positively charged and silk becomes n

charged.

Electrons in fur are loosely bound in it than the electrons in ebon



It is very important to note that the electrification of the body (whe

positive or negative) is due to transfer of electrons from one body

i.e. If the electrons are transferred from a body, then the deficiencelectrons makes the body positive.

If the electrons are gained by a body, then the excess of electrons

body negative.

If the two bodies from the following list are rubbed, then the body

early in the list is positively charges whereas the latter is negative

Fur, Glass, Silk, Human body, Cotton, Wood, Sealing wax, Amber

Sulphur, Rubber, Ebonite.

Amber, Ebonite, Rubber, PlasWool, Flannel

SilkGlass

Column II (-ve Charge)Column I (+ve Charge)



Properties of Charges:

1. There exists only two types of charges, namely positive and ne

2. Like charges repel and unlike charges attract each other.

3. Charge is a scalar quantity.

4. Charge is additive in nature. eg. +2 C + 5 C – 3 C = +

5. Charge is quantized.

i.e. Electric charge exists in discrete packets rather than in con

amount.

It can be expressed in integral multiples fundamental electroni

(e = 1.6 x 10-19 C)

q = ± ne where n = 1, 2, 3, …………

6. Charge is conserved.

i.e. The algebraic sum of positive and negative charges in an issystem remains constant.



Note: Recently, the existence of quarks of charge ⅓ e and ⅔ e hapostulated. If the quarks are detected in any experiment with con

practical evidence, then the minimum value of ‘quantum of chargeeither ⅓ e or ⅔ e. However, the law of quantization will hold good

Coulomb’s Law – Force between two point electric ch

The electrostatic force of interaction (attraction or repulsion) betw

electric charges is directly proportional to the product of the charproportional to the square of the distance between them and actsjoining the two charges.

Strictly speaking, Coulomb’s law applies to stationary point charg

r

q1F α q1 q2

F α 1 / r2

or F αq

1

q2

r2F = k

q1

q2

r2or where k is a positiveproportionality called

l i f



In medium, k =1

4πεwhere ε is the absolute electric pe

the dielectric medium

The dielectric constant or relative permittivity or specific inductiv

dielectric coefficient is given by

F =q1 q2

r2

1

4πε0

In vacuum,

F =

q1 q2

r2

1

4πε0εr

In medium,

ε0 = 8.8542 x 10-12 C2 N-1 m-2

K = εr =εε0



Coulomb’s Law in Vector Form:

r

+ q1

F12

r12

q1q2 > 0

+ q1 r1

In vacuum, for q1 q2 > 0,

q1 q2

r2

1

4πε0

r21F12 =

q1 q2

r2

1

4πε0

r12F21 =

In vacuum, for q1 q2 < 0,

q1 q21rF

q1 q21rF&

r

- q1

F12

r12

q1q2 > 0



q1 q2

r3

1

4πε0

r12F12 =

q1 q2

r3

1

4πε0

r2F21 =&

Note: The cube term of the distance is simply because of vector

Otherwise the law is ‘Inverse Square Law’ only.

Units of Charge:

In SI system, the unit of charge is coulomb (C).

One coulomb of charge is that charge which when placed at rest idistance of one metre from an equal and similar stationary chargeis repelled by it with a force of 9 x 109 newton.

In cgs electrostatic system, the unit of charge is ‘statcoulomb’ or

In cgs electrostatic system, k = 1 / K where K is ‘dielectric consta

For vacuum, K = 1.

F =q

1

q2

r2



1 emu of charge = c esu of charge

1 emu of charge = 3 x 1010 esu of charge

1 coulomb of charge = 3 x 109 statcoulomb

1 abcoulomb = 10 coulomb

Relative Permittivity or Dielectric Constant or SpecificCapacity or Dielectric Coefficient:

The dielectric constant or relative permittivity or specific inductive

dielectric coefficient is given by the ratio of the absolute permittivmedium to the permittivity of free space.

K = εr =εε0

The dielectric constant or relative permittivity or specific inductive

dielectric coefficient can also be defined as the ratio of the electrobetween two charges separated by a certain distance in vacuum to



Continuous Charge Distribution:Any charge which covers a space with dimensions much less than

away from an observation point can be considered a point chargeA system of closely spaced charges is said to form a continuous c

distribution.

It is useful to consider the density of a charge distribution as we d

of solid, liquid, gas, etc.

(i) Line or Linear Charge Density ( λ ):

If the charge is distributed over a straight line or over the circumfecircle or over the edge of a cuboid, etc, then the distribution is cal

charge distribution’.

Linear charge density is the charge per unit length. Its SI unit is C

qλ =

lλ =

dq

dl

or

+ + + + + + +

dq

dl



(ii) Surface Charge Density ( σ ):

σ =q

Sσ =

dq

dSor

If the charge is distributed over a surface area, then the distributio

‘surface charge distribution’.

Surface charge density is the charge per unit area. Its SI unit is C

+ + + + + + + +

+ + + + + + + ++ + + + + + + ++ + + + + + + +

dq

dSTotal charge on surface S, q = ∫ σ dS

S

(iii) Volume Charge Density ( ρ ):

ρ =q

ρ =dq

or

If the charge is distributed over a volume, then the distribution is c‘volume charge distribution’.

Volume charge density is the charge per unit volume. Its SI unit is



Electric Field:Electric field is a region of space around a charge or a system of

within which other charged particles experience electrostatic forcTheoretically, electric field extends upto infinity but practically it i

certain distance.

Electric Field Strength or Electric Field Intensity or El

Electric field strength at a point in an electric field is the electrostunit positive charge acting on a vanishingly small positive test chat that point.

E =∆qFLt

∆q → 0 q0

FE =or or 14πε0

E =

+ q0+ q - q

q – Source charge, q0 – Test charge, F – Force & E

FF



Note:

1. Since q0 is taken positive, the direction of electric field ( E ) isdirection of electrostatic force ( F ).

2. Electrostatic force on a negatively charged particle will be opdirection of electric field.

3. Electric field is a vector quantity whose magnitude and direcuniquely determined at every point in the field.

4. SI unit of electric field is newton / coulomb ( N C-1 ).



Electric Field due to a Point Charge:

O

Z

Y

+

+ q

r

Force exerted on q0 by q is

q q0

r2

1

4πε0

rF =

q q0

r3

1

4πε0

rF =

Electric field strength isq0

FE =

q

r31

4πε0

E (r) = r

or

orr2

1

4πε0

E (r) =q

r

Th l t i fi ld d t i t h h

E



Electric field in terms of co-ordinates is given by

( x2 + y2 + z2 ) 3/2

1

4πε0E (r) =

q

i j k( x + y + z )

Superposition Principle:

- q5

+ q4

+ q1

F12

F15

The electrostatic force experienced by acharge due to other charges is the vector

sum of electrostatic forces due to theseother charges as if they are existingindividually.

F1 = F12 + F13 + F14 + F15

F12

F

qa

qb

1

4πε0

Fa (ra) = ∑b=1b≠a

Nra - rb

ra - rb│ │3



Superposition principle holds good for electric field also.

Note:

The interactions must be on the charge which is to be studied duecharges.

The charge on which the influence due to other charges is to be f

assumed to be floating charge and others are rigidly fixed.

For eg. 1st charge (floating) is repelled away by q2

and q4

and attra

q3 and q5.

The interactions between the other charges (among themselves)

ignored. i.e. F23, F24, F25, F34, F35 and F45 are ignored.

Electric Lines of Force:

An electric line of force is an imaginary straight or curved path alounit positive charge is supposed to move when free to do so in an

field.Electric lines of force do not physically exist but they represent re



1. Electric Lines of Force due to a Point Charge:

q < 0q > 0

a) Reofinfie

Tharre

stel

b) Re

of

infie



2. Electric Lines of Force due to apair of Equal and Unlike Charges:

(Dipole)

+ q

- q

P

E

3. Electric Lines of Fopair of Equal and Li

.N



4. Electric Lines of Force due to a Uniform Field:

+

+++

Properties of Electric Lines of Forceor Field Lines:

1. The electric lines of force are imaginary lines.

2. A unit positive charge placed in the electric field tends to followalong the field line if it is free to do so.

3. The electric lines of force emanate from a positive charge and t

a negative charge.

4. The tangent to an electric field line at any point

gives the direction of the electric field at that point.

5. Two electric lines of force can never crosseach other. If they do, then at the point ofintersection, there will be two tangents. It

means there are two values of the electric fieldat that point, which is not possible. E

+1 C



6. Electric lines of force are closer

(crowded) where the electric field

is stronger and the lines spreadout where the electric field isweaker.

7. Electric lines of force are

perpendicular to the surface of a

positively or negatively chargedbody.

Q > q

Q

8 Electric lines of force contract lengthwise to represent attractio



10.The number of lines per unit cross – sectional area perpendicu

field lines (i.e. density of lines of force) is directly proportional

magnitude of the intensity of electric field in that region.

11. Electric lines of force do not pass through a conductor. Henceof the conductor is free from the influence of the electric field.

Solid or hollowconductor

No Field

+

++

++

+

+

----

--

-

++++

E E

----

α E∆N

∆A

(Electrostatic



Electric Dipole:

Electric dipole is a pair of equal and opposite charges separated small distance.

The electric field produced by a dipole is known as dipole field.

Electric dipole moment is a vector quantity used to measure the s

electric dipole.

- q2 l

p

p = (q x 2l) l

The magnitude of electric dipole moment is the product of magnit

charge and the distance between the two charges.

The direction is from negative to positive charge.

The SI unit of ‘p’ is ‘coulomb metre (C m)’.

Note:



Electric Field Intensity due to an Electric Dipole:

i) At a point on the axial line:

Resultant electric field intensityat the point P is

EP = EA + EB

The vectors EA and EB are

collinear and opposite.

1

4πε0

iEA =q

(x + l)2

q

(x - l)2

1

4πε0

iEB =

│EP │ = │EB│ - │EA│

│EP │ = q

( l)2

q

( l)2

1

4πε ][ -

│EP │ =1

4πε0 (

If l << x, thenEP

ll x

+ q- qp

A B E

O

EP =1

4πε0

2

(x2



θθ

ll

y

θ

θ

+ q- qp

A B

EB

EA

EQQ

Resultant electric field intensity

at the point Q is

EQ = EA + EB

The vectors EA and EB are

acting at an angle 2θ.

ii) At a point on the equatorial line:

q

( x2 + l2 )

1

4πε0

EA = i

q1

4πε0EB = i( x2 + l2 )

E

EA co

EB co

The vectors EA sin θ and EB sin θare opposite to each other andhence cancel out.

The vectors EA cos θ and EB cos θ

EQ =q2

4πε0 ( x2 + l2 ) ( x

1E

q . 2l

O



EQ =

1

4πε0

p

( x2 + l2 )3/2 (- i )

If l << y, then

EQ ≈ p

4πε0 y3

The direction of electric field intensity at a point on the equatorial

dipole is parallel and opposite to the direction of the dipole mome

If the observation point is far away or when the dipole is very sho

electric field intensity at a point on the axial line is double the elecintensity at a point on the equatorial line.



Torque on an Electric Dipole in a Uniform Electric Fie

The forces of magnitude pE act

opposite to each other and hence netforce acting on the dipole due toexternal uniform electric field is zero.So, there is no translational motion ofthe dipole.

θ

However the forces are along different

lines of action and constitute a couple.Hence the dipole will rotate andexperience torque.

Torque = Electric Force x distance

θ

t = q E (2l sin θ)

= p E sin θ

q E- q

2l

t

p

Case i: If θ = 0°, thet = p x E

p



Work done on an Electric Dipole in Uniform Electric Fi

dW = tdθ

= p E sin θ dθ

W = ∫ p E sin θ dθ

W = p E (cosθ1 - cos θ2)

θ1

θ2

If Potential Energy is arbitrarily taken zero when the dipole is at 9then P.E in rotating the dipole and inclining it at an angle θ is

Potential Energy U = - p E cos θ

Note: Potential Energy can be taken zero arbitrarily at any positidipole.

- q

2l θ

q E

q E

When an electric dipole is placed in a uniform electric field, it exp

torque and tends to allign in such a way to attain stable equilibriu

d



Line Integral of Electric Field (Work Done by Electric F

Negative Line Integral of Electric Field represents the work done b

field on a unit positive charge in moving it from one point to anotelectric field.

O

Z

Y

+ q

ArA

WAB = dW = - E . dl

A

B

Let q0 be the test charge in place of the unitpositive charge.

The force F = +q0E acts on the test charge

due to the source charge +q.

It is radially outward and tends to acceleratethe test charge. To prevent thisacceleration, equal and opposite force –q0Ehas to be applied on the test charge.

Total work done by the electric field on the test charge in movingin the electric field is



Electric Potential

Review



Electric potential is a physical quantity which determines the flow

from one body to another.

It is a physical quantity that determines the degree of electrificatio

Electric Potential at a point in the electric field is defined as the w

moving (without any acceleration) a unit positive charge from infi

point against the electrostatic force irrespective of the path follow

Electric Potential:

=11qq0

4πε0

][ -rB rA

WAB = - E . dl

A

B

=q

4πε0

WAB

q0

According to definition, rA = ∞ and rB = r

(where r is the distance from the sourcand the point of consideration)

=q

4

W∞B

q= V V =

W∞B

q

or



Electric Potential due to a Single Point Charge:

r

+ q B

x

Q

dxE +q0

Let +q0 be the test charge

placed at P at a distance xfrom the source charge +q.

To prevent this acceleration, equal and opposite force –q0E has to

on the test charge.

The force F = +q0E is

radially outward and tends

to accelerate the test charge.

Work done to move q0 from P to Q through ‘dx’ against q0E is

dW = F . dx = q0E . dx dW = q0E dx cos 180° = - q 0E

dW = - dxq q0

4πε0 x2E =

q

4πε0 x2

or

Total work done to move q0 from A to B (from ∞ to r ) is W∞q0

P



Electric Potential due to a Group of Point Charges:

+qn

q4

r4

rnVP = V1 + V2 + V3 + V4 + …………+ Vn

│ │

1

4πε0V = ∑

i=1

n qi

r - ri

( in terms of

position vector )

The net electrostatic potential at a point in the

electric field due to a group of charges is thealgebraic sum of their individual potentials at thatpoint.

1. Electric potential at a point due to a charge is not affected by tof other charges.

2. Potential, V α 1 / r whereas Coulomb’s force F α 1 / r2.

1

4πε0

V = ∑i=1

n qi

ri



Electric Potential due to an Electric Dipole:

llx

+ q- qp

A B

qVP q+

=4πε0 (x – l)

1

VP = VP q++ VP q-

VP q-=

4πε0 (x + l)1

- q

VP =4πε0

q

(x – l)

1[ -

(x + l)

1

]

VP =1

4πε0

q . 2l

(x2 – l2)

i) At a point on the axial line:

O



- qp

A θθ

y

O

ii) At a point on the equatorial line:

qVQ q+=

4πε0 BQ1

VQ = VP q++ VP q-

VQ q-=

4

πε0AQ

1 - q

VQ =4πε0

q

BQ

1[ -AQ

1 ]

VQ = 0 BQ = AQ

Q

ll



Equipotential Surfaces:

A surface at every point of which the potential due to charge distr

the same is called equipotential surface.

i) For a uniform electric field:

E

V1 V2V3

Plane Equipotential Surfaces +



Properties of Equipotential Surfaces:

2. The electric field is always perpendicular to the element dl of th

equipotential surface.

1. No work is done in moving a test charge from one point to anot

equipotential surface.

WAB

q0

VB - VA = ∆V =

If A and B are two points on the equipotential surface, then VB

WAB

q0= 0 or WAB

= 0

WAB

= -E . dl

A

B

= 0

Since no work is done on equipotential surface,

i.e.E dl cos θ =



ii) Electrostatic Potential Energyof a Three Charges System:

OZ

Y

A (q1)

r1

or

r

U =q1q2

4πε0 │ │- r1r2

1+

q1q3

4πε0 │ │- r1r3

1

+

q2q3

4πε0 │ │- r2r3

1

U =

q1q2

4πε0 r12

1

[

q1q3

r31

q2q3

r32+ + ]iii) Electrostatic Potential Energy of an n - Charges Syste

]



Electric Flux &

Gaussian Theorem

Review



Area Vector:

Small area of a surface can be represented by a vector.

SElectric Flux:

Electric flux linked with any surface is defined as the total numbe

lines of force that normally pass through that surface.

90°

θ

dS

S

Electric flux dΦ through a small area

element dS due to an electric field E at an

angle θ with dS is

= E dS cos θE . dSdΦ =

Total electric flux Φ over the wholesurface S due to an electric field E is

Φ = E . dS = E S cos θ = E . S

dS = dS n

dS



Solid Angle:

Solid angle is the three-dimensional equivalent of an ordinary two

dimensional plane angle.

SI unit of solid angle is steradian.

Solid angle subtended by area element dS at thecentre O of a sphere of radius r is

d =dS cos θ

r2

dS θ

1. For 0°< θ < 90°, Φ is positive.

2. For θ = 90°, Φ is zero.

3. For 90°< θ < 180°, Φ is negative.

Special Cases:

r



Gauss’s Theorem:

The surface integral of the electric field intensity over any closed

surface (called Gaussian surface) in free space is equal to 1 / ε0 ticharge enclosed within the surface.

E . dS =

S

ΦE =1

ε0

∑i=1

n

qi

Proof of Gauss’s Theorem for Spherically Symmetric

E . dSdΦ =r2

1

4πε0

=q

r . dS n

dΦ =r2

1

4πε0

q dSr n.

Here, = 1 x 1 cos 0°= 1r n.

dΦ1 q dS

O •

r

r+q



Proof of Gauss’s Theorem for a Closed Surface of any

E . dSdΦ =r2

14πε0

= q r . dS n

dΦ =

r2

1

4πε0

q dSr n.

Here, = 1 x 1 cos θ= cos θ

r n.

dΦ =r2

q

4πε0

dS cos θ

Φ dΦq

4q

dq

d

r

+q •



O •

r

r+q

Deduction of Coulomb’s Law from Gauss’s Theorem:

From Gauss’s law,

E . dS =

S

ΦE =q

ε0

E dS =

S

ΦE =q

ε0

or dS =

S

ΦE =

q

ε0E

E =q

4πε0 r2E x 4π r2

q

ε0

=

If a charge q0 is placed at a point where E

Since E and dS are in the same direction,

or



Applications of Gauss’s Theorem:

1. Electric Field Intensity due to an Infinitely Long Straight

Wire:

Gaussia

closed saround distribu

that theintensit

fixed vapoint on

From Gauss’s law,

E . dS =

S

ΦE =q

ε0

E . dS = E . dS + E . dS + E . dS

- ∞ B A

C

E

E E

dSdS

dS

l

r



q

ε0

=λ l

ε0(where λ is the liner charge density)

E x 2 π r l =λ l

ε0

or E = 2 πε0

1 λ

r

or E =4 πε0

1 2λr

In vector form, E (r) =4 πε0

1 2λr

r

The direction of the electric field intensity is radially outward from

line charge. For negative line charge, it will be radially inward.



dS

C

BA

E dS r

l

2. Electric Field Intensity due to an Infinitely Long, Thin PlaCharge:

From Gauss’s law,

σ

E . dS =

S

ΦE =q

ε0

E . dS = E . dS + E . dS + E . dS

TIP:

The fieldstraight,

uniforml



(where σ is the surface charge density)

or E =2 ε0

σ

In vector form, E (l) = 2 ε0

σ

l

The direction of the electric field intensity is normal to the plane a

from the positive charge distribution. For negative charge distrib

be towards the plane.Note:

The electric field intensity is independent of the size of the Gauss

constructed. It neither depends on the distance of point of considthe radius of the cylindrical surface.

q

ε0

=σ π r2

ε0

2 E x π r2 =σ π r2

ε0

If the plane sheet is thick, then the charge distribution will be ava



3. Electric Field Intensity due to Two Parallel, Infinitely LonPlane Sheet of Charge:

σ1σ2

E1E1

E2E2E2

E E E

Region I Region II Reg

σ1 > σ2( )

Case 1: σ1 > σ2



Case 2:

σ1 σ2

E1E1

E2E2E2

E E E


(σ1 > σ2

( )

+ σ1 & - σ2



Case 3:

σ1 σ2

E1E1

E2E2E2

E = 0 E ≠ 0


E

E E

+ σ & - σ



4. Electric Field Intensity due to a Uniformed Charged This Shell:

q

r

R

From Gauss’s law,

E . dS =

S

ΦE =q

ε0

E dS =

S

ΦE =q

ε0

or dS =

S

ΦE =q

ε0

E

E x 4π r2 q=


or E = q

4 2

i) At a point P outside the shell:

Electric field due charged thin sphe

HOLLOW

……… Gaussia

O •



From Gauss’s law,

E . dS =

S

ΦE =q

ε0

E dS =

S

ΦE =q

ε0

or dS =

S

ΦE =q

ε0

E

E x 4π R2q

ε0

=


or E =q

4πε0 R2

ii) At a point A on the surface of the shell:

Electric field due Since q = σ x 4π R2

q

HOLLOW

O •



r’

Oq

HOLLO

•

From Gauss’s law,

E . dS =

S

ΦE =q

ε0

E dS =

S

ΦE =q

ε0

or dS =

S

ΦE =q

ε0

E

E x 4π r’2q

ε0=


or E =0

4πε0 r’2

iii) At a point B inside the shell:

(since q = 0 inside the Gaussian surface)

E

Emax



Capacitance

Review



Behaviour of Conductors in the Electrostatic Field:

1. Net electric field intensity in the interior of a

conductor is zero.When a conductor is placed in an

electrostatic field, the charges (freeelectrons) drift towards the positive plateleaving the + ve core behind. At an

equilibrium, the electric field due to thepolarisation becomes equal to the appliedfield. So, the net electrostatic field insidethe conductor is zero.

E

Ene

2. Electric field just outside the chargedconductor is perpendicular to the surface ofthe conductor.

Suppose the electric field is acting at an

angle other than 90°, then there will be a

component E cos θ acting along the tangentat that point to the surface which will tend to

E

T P



3. Net charge in the interior of a conductor is zero.

The charges are temporarily separated. The total

charge of the system is zero.

E . dS =

S

ΦE =q

ε0

Since E = 0 in the interior of the conductor,

therefore q = 0.

4. Charge always resides on the surface of a

conductor.

Suppose a conductor is given some excess

charge q. Construct a Gaussian surface justinside the conductor.

Since E = 0 in the interior of the conductor,therefore q = 0 inside the conductor.

q

5. Electric potential is constant for the entireconductor



6. Surface charge distribution may be different

at different points.

σ=

q

S σmin

Every conductor is an equipotential volume

(three- dimensional) rather than just anequipotential surface (two- dimensional).

Electrical Capacitance:

The measure of the ability of a conductor to store charges is kn

capacitance or capacity (old name).

q α V or q = C V or C =

q

VIf V = 1 volt, then C = q

Capacitance of a conductor is defined as the charge required to

potential through one unit.

SI Unit of capacitance is ‘farad’ (F). Symbol of capacitance:



Capacitance of an Isolated Spherical Conductor:

Let a charge q be given to the sphere which

is assumed to be concentrated at the centre.

Potential at any point on the surface is

V =q

4πε0 r

C =q

V

C = 4πε0 r

1. Capacitance of a spherical conductor is directly proportional t

2. The above equation is true for conducting spheres, hollow or s

3. IF the sphere is in a medium, then C = 4πε0εr r.



Principle of Capacitance:A

A

Step 1: Plate A is positively charged and B is neutral.

Step 2: When a neutral plate B is brought near A,charges are induced on B such that the side near A is

negative and the other side is positive.

The potential of the system of A and B in step 1 and 2

remains the same because the potential due to positiveand negative charges on B cancel out.

Step 3: When the farther side of B is earthed thepositive charges on B get neutralised and B is left onlywith negative charges.

Now, the net potential of the system decreases due tothe sum of positive potential on A and negativepotential on B.

To increase the potential to the same value as was in

step 2, an additional amount of charges can be given tol t A

P



Capacitance of Parallel Plate Capacitor:

Parallel plate capacitor is an arrangement of two

parallel conducting plates of equal areaseparated by air medium or any other insulating

medium such as paper, mica, glass, wood,ceramic, etc.

Aσ

V = E dσε0

= d

orq d

V =A ε0

But dC =

A ε0

C =

q

V

If the space between the plates is filled with dielectric medium of permittivity εr, then

dC =

A ε0 εr



Series Combination of Capacitors:

V1 V2

V

C1 C2In series combination,

i) Charge is same in each capacitor

ii) Potential is distributed in inverse

proportion to capacitances

i.e. V = V1

+ V2

+ V3

Butq

V1 =C1

V2 =C2

qV3 =

C3

q, and

qV =

C

,

(where C is the equivalent cap

effective capacitance or net catotal capacitance)

q

= C1+ C2

q

+ C3

qq

C

or

q q

1

C=

1=

C1

+C2

1+

C3

11

C



Parallel Combination of Capacitors:

In parallel combination,

i) Potential is same across each capacitor

ii) Charge is distributed in direct proportion to

capacitances

i.e. q = q1 + q2 + q3

But , and,

(where C is the equivalent

capacitance)

or

Th ff ti it i th f th i di id l it

q1 = C1 V q2 = C2 V q3 = C3 V q = C V

C V = C1V + C2 V + C3 V

∑i=1

n

CiC =C = C1 + C2 + C3



Energy Stored in a Capacitor:

The process of charging a capacitor is

equivalent to transferring charges from oneplate to the other of the capacitor.

The moment charging starts, there is a potentialdifference between the plates. Therefore, totransfer charges against the potential difference

some work is to be done. This work is stored aselectrostatic potential energy in the capacitor.

If dq be the charge transferred against the

potential difference V, then work done is

dU = dW = V dq

q=

Cdq

The total work done ( energy) to transfer charge q is



Energy Density:

U =

1

2C V2

dC =

A ε0 V = E dand

U =1

2ε0 Ad E2

1

2ε0 E2=

U

Ad

1

2ε0 E2=U

But

or or

SI unit of energy density is J m-3.

Energy density is generalised as energy per unit volume of the fi

Energy Stored in a Series Combination of Capacitors:

1=

C1

+C2

1+

C3

11

C+

Cn

1………. +

U =q21 U = 1

2q2 1[ C

+C1 +

C1 +

C1………. +



Energy Stored in a Parallel Combination of Capacitors

U = U1 + U2 + U3 + ………. + Un

The total energy stored in the system is the sum of energy stored

individual capacitors.

C = C1 + C2 + C3 + ……….. + Cn

U =1

2C V2 U =

1

2V2 ( C1 + C2 + C3 + ……….. + Cn )

Loss of Energy on Sharing of Charges between the Ca

Parallel:Consider two capacitors of capacitances C1, C2, charges q1, q2 anpotentials V1,V2.

Total charge after sharing = Total charge before sharing

(C + C ) V = C V + C V



The total energy before sharing is

Ui =

1

2C1 V12

1

2C2 V22+

The total energy after sharing is

Uf =

1

2(C1 + C2) V2

Ui – Uf =C1 C2 (V1 – V2)2

2 (C1 + C2)

Ui – Uf > 0 or Ui > Uf

Therefore, there is some loss of energy when two charged capacit

connected together.



Polar Molecules:A molecule in which the centre of positive charges does

not coincide with the centre of negative charges is calleda polar molecule.

Polar molecule does not have symmetrical shape.

Eg. H Cl, H2 O, N H3, C O2, alcohol, etc.H

Effect of Electric Field on Polar Molecules:E = 0 E

p = 0 p

In the absence of external electricfield the permanent dipoles of the

When electric field is adipoles orient themse



Non - polar Molecules:

A molecule in which the centre of positive charges coincides wit

negative charges is called a non-polar molecule.Non-polar molecule has symmetrical shape.

Eg. N2 , C H4, O2, C6 H6, etc.

Effect of Electric Field on Non-polar Molecules:E = 0 E

p = 0 p

In the absence of externalelectric field the effective

When electric field is applied, thecharges are pushed in the direct



Dielectrics:Generally, a non-conducting medium or insulator is called a ‘diele

Precisely, the non-conducting materials in which induced chargeson their faces on the application of electric fields are called dielect

Eg. Air, H2, glass, mica, paraffin wax, transformer oil, etc.

Polarization of Dielectrics:

E0

When a non-polar dielectric slab issubjected to an electric field, dipoles

are induced due to separation ofeffective positive and negative centres.

E0 is the applied field and Ep is theinduced field in the dielectric.

The net field is EN = E0 – Ep

i.e. the field is reduced when a

dielectric slab is introduced.

E = 0



Capacitance of Parallel Plate Capacitor with Dielectric

EpE0 EN = E0 -

V = E0

(d – t) + EN

t

EN

K =E0 or EN =

K

E0

V = E0 (d – t) +KE0 t

V = E0 [ (d – t) +K

t]

But E0 =ε0

σ=

ε0

qA

and C =q

VA ε

or C =

A ε0

d [1 –K

t

d

t(1 -

or C =

C0

[ tt(



If the dielectric slab occupies the whole space between the plates,

then

WITH DIELECTRIC SLAB

Remains the sDecreases

EN = E0 – Ep

Electric Field

Increases (K Increases (K C0)Capacitance

Increases (K Remains the sameCharge

With Batteryconnected

With Batterydisconnected

Physcial Quantity

C0

K =C

C = K C0

Dielectric Constant



Van de Graaff Generator:

C2

C1

P2

S

S – Large Copper sphere

C1, C2 – Combs with sharp points

P1, P2 – Pulleys to run belt

HVR – High Voltage Rectifier

M – Motor

IS – Insulating Stand

D – Gas Discharge Tube

T - Target



Principle:

Consider two charged conducting spherical shells such that osmaller and the other is larger. When the smaller one is kept inlarger one and connected together, charge from the smaller ontransferred to larger shell irrespective of the higher potential o

shell. i.e. The charge resides on the outer surface of the outer

the potential of the outer shell increases considerably.

Sharp pointed surfaces of a conductor have large surface char

densities and hence the electric field created by them is very hcompared to the dielectric strength of the dielectric (air).

Therefore air surrounding these conductors get ionized and th

charges are repelled by the charged pointed conductors causi

discharging action known as Corona Discharge or Action of Posprayed charges moving with high speed cause electric wind.

Opposite charges are induced on the teeth of collecting comb



Construction:

Van de Graaff Generator consists of a large (about a few mradius) copper spherical shell (S) supported on an insulating s

which is of several metres high above the ground.

A belt made of insulating fabric (silk, rubber, etc.) is made tthe pulleys (P1, P2 ) operated by an electric motor (M) such that

on the side of the combs.

Comb (C1) near the lower pulley is connected to High Volta

(HVR) whose other end is earthed. Comb (C2) near the upper pconnected to the sphere S through a conducting rod.

A tube (T) with the charged particles to be accelerated at it

the target at the bottom is placed as shown in the figure. The b

of the tube is earthed for maintaining lower potential.



Working:

Let the positive terminal of the High Voltage Rectifier (HVR) iconnected to the comb (C1). Due to action of points, electriccaused and the positive charges are sprayed on to the belt (srubber). The belt made ascending by electric motor (EM) and

(P1) carries these charges in the upward direction.

The comb (C2) is induced with the negative charges which arcarried by conduction to inner surface of the collecting sphe(dome) S through a metallic wire which in turn induces posi

charges on the outer surface of the dome.

The comb (C2) being negatively charged causes electric windspraying negative charges due to action of points which neuthe positive charges on the belt. Therefore the belt does not

any charge back while descending. (Thus the principle ofconservation of charge is obeyed.)



The process continues for a longer time to store more and mcharges on the sphere and the potential of the sphere increa

considerably. When the charge on the sphere is very high, t

leakage of charges due to ionization of surrounding air alsoincreases.

Maximum potential occurs when the rate of charge carried in

the belt is equal to the rate at which charge leaks from the shdue to ionization of air.

Now, if the positively charged particles which are to be

accelerated are kept at the top of the tube T, they get accelerdue to difference in potential (the lower end of the tube isconnected to the earth and hence at the lower potential) and

made to hit the target for causing nuclear reactions, etc.



Uses:

Van de Graaff Generator is used to produce very highpotential difference (of the order of several million volts) fo

accelerating charged particles.

The beam of accelerated charged particles are used totrigger nuclear reactions.

The beam is used to break atoms for various experimein Physics.

In medicine, such beams are used to treat cancer.

It is used for research purposes.

Electrics Review

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