AZAD SLEMAN HSEN AZAD SLEMAN HSEN

ELECTRYCITY AND MAGNETIC

ELECTRYCITY AND MAGNETIC

COLLEGE OF SCINCE COLLEGE OF SCINCE

physic department physic department

2011 2011

قال الله:

•Electric field

•Electric potential

•Capcitor

•Current

•Resistance

•Electric field

•Electric potential

•Capcitor

•Current

•Resistance

ده مي كو باركه جهي خو دكوريت ده مي كو باركه جهي خو دكوريت ئه فه به يدادبن ئه فه به يدادبن

POSITIVE charge moving in an E field.

(a) When a positive charge moves in the direction of an electric field, the field does positive work and the potential energy decreases. Work = qo E d

The charge qo moves along a straight

line extending radially from charge q.

Electric field E and potential V

at points inside and outside a positively charged spherical

conductor.See

www.physics.sjsu.edu/becker/physics51

Coulomb’s law:

EpW

dqEW

.

.

Electric field:

Work = Force x distance

204

1

r

qE

r

qrV

04

1)(

WdFPEdqEPEPE ..12

r

QQrU

21

4

1)(

0

SI Unit of electric potential: Volt (V)

electric potential=potential Energy/charge

dEq

dqEV .

.

JMNW ,.

2/12200 )(4

1

4

1

dx

dx

r

dqdV

dxdq

A rod of length L located along the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin.

ddLL

dxx

dxxdx

dxdVV

L

LL

ln)(ln4

)(ln

)(2

1.2ln

4)(4

2/122

0

02/122

0

2/1122

002/122

0

d

dLLV

2/122

0

)(ln

4

xdr

azad

P

R r

xx

Every dQ of charge on the ring is the same distance from the point P.

2 2

dq dqdV k k

r x R

2 2ring ring

dqV dV k

x R

2 2

kQV

x R

B) Use Ex = - dV/dx Ex = - (x2 + a2 )dV -1/2

kQx

dx= - kQ

d

dx

Ex =

= - kQ (- ½)

(x2 + a2 )3/2

2x (x2 + a2 )-3/2 What about Ey and Ez ?

What is the electric potential at the center of the ring?

What is the electric field at the center of the ring?

Example: A disc of radius R has a uniform charge per unit area and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center.

P

r

xxR

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).We *can use the equation

for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.

R

2 2 2 2ring ring 00 0

1 2 rdr rdrV dV

4 2x r x r

R

2 2 2 2 2 22

0 0 00

QV x r x R x x R x

2 2 2 R

2

Q

R

2 22

0

QV x R x

2 R

rx

dQ

Remembe (r+x).r its uniform with intgral in content mrs azad

Note:in the z direction you change z place

X ok and you can countinio on the exampl:

(B) Find the electric potential at point P located on the perpendicular central axis of a uniformly charged disk of radius a and surface charge density . (b) Find an expression for the magnitude of the electric field at point P. (b) Ex = - dV/dx

k (x2 + a2 )1/2 - x

Ex = -dV

dx

= - d

dx

x2 + a2

x 1 -Ex = k

x2 + a2

x 1 -k

When you are really close to this disk, then it is as if you are looking at an infinite plane of charge, use above equation to deduce the electric field. Is the result consistent with the result obtained from our discussion using Gauss’s law ?

Electric Potential due to uniformly charged annulus

Calculate the electric potential at point P on the axis of an annulus, which has uniform charge density .

drrdAdq

xr

dqkdV e

2 where

22

V= 2k [ (x2+b2)1/2 - (x2+a2)1/2]

b

a

exr

drrkV

222

Ok you can restory this function

Electric Potential due to non-uniformly charged disk

A disk of radius R has a non-uniform surface charge

density =Cr where C is constant and r is distance from the center of the disk as shown. Find the potential at P.

dV

kedq

r2 x2

dq = dA = Cr(2πrdr) and

R

exr

drrkCV

022

2

)2(

Standard

integral

V= C(2k) {R(R2+x2)1/2 +x2ln(x/[R+ (R2+x2)1/2])}

r

Electric Potential due to a finite line of chargeA rod of length 2 l located along the x axis has a total charge

Q and a uniform linear charge density =Q/2l . Find the electric potential at a point P located on the y axis a distance a from the origin.

l

V = k

dxl

x2 + a2

- l

V =kQ

2 lln

l 2+a2 + l

l 2+a2 - l

How is this result consistent with the E field for infinite line of charge obtained using Gauss’s Law? (Homework)

Electric Potential due to a finite line of chargeA rod of length L as shown lies along the x axis with its left end at the origin and has a non-uniform charge density = x (where is a positive constant). (a) What are the units of ? (b) Calculate the electric potential at point A. (c) Calculate the electric potential at point B.(a) C/m2

V = k rdq = k r

dx

L

xd

dxxk

0)(

= k [ L – d ln( 1 + L/d) ]

Electric Potential due to a uniformly charged sphere

An insulating solid sphere of radius R has a uniform positive volume charge density and total charge Q.

(a) Find the electric potential at a point outside the sphere, that is, r > R. Take the potential to be zero at r = .

(b)Find the potential of a point inside the sphere, (r < R). Outside the sphere, we have Er =

k Q

r2For r > R

To obtain potential at B, we use

VB = -r

Er dr = - kQ

r

r2

drVB = k Q

r

Potential must be continuous at r = R, => potential at surface

VC = k Q

R

Inside the sphere, we have k Q Er =

R3For r < R

To obtain the potential difference at D, we use

VD - VC = -r

Er dr = - r dr =

r

R

k Q

R3

r

R

k Q

2R3 ( R2 – r2 )

VC = k Q

RSince

To obtain the absolute value of the potential at D, we add

the potential at C to the potential difference VD -

VC :VD = k Q

2R3 -

r2

R2

Check V for r = R

For r < R

Outside the sphere, we have

k Q Er = r2

For r > R

To obtain potential at B, we use

VB = -r

Er dr = - kQ

r

r2

dr

VB = k Q

r

Electric Potential due to a uniformly charged sphere

Pictures from Serway & Beichner

Inside the sphere, we have k Q

Er = R3For r < R

To obtain the potential difference at D, we use

VD - VC = -r

Er dr = - r dr =

r

R

k Q

R3

r

R

k Q

2R3 ( R2 – r2 )

VC = k Q

RSince

To obtain the absolute value of the potential at D, we add

the potential at C to the potential difference VD - VC :VD = k Q2R

3 - r2

R2

Check V for r = R

For r < R

Example: Two connected Charged Conducting Spheres

Two spherical conductors of radii r1

and r2 are separated by a distance

much greater than the radius of either sphere. The spheres are connected by a conducting wire. The charges on the spheres in equilibrium (the spheres are at the SAME electric potential V) are q1 and q2

respectively, the they are uniformly charged. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres. Connected both must have the same electric

potential. V = kq1/r1 = kq2/r2 q1/r1= q2/r2

Sphere far apart charge uniform magnitude of surface E given by

E1 = kq1/r12 and E2 = kq2/r2

2

we have E1/E2 = r2/r1

L

P

a

0

a

La

x

dxλdQ

x

QkV

dd

La

a x

xkdVV

d

La

axk

ln

aLak lnln

a

La ln

x

A rod of length L located along the X axis has a uniform linear charge density λ. Find the electric potential at a point P located on the X axis a distance a from the origin.

x

dxkV

d

Electric Potential due to uniformly charged ring

Example 25 Q41 here

Consider a ring of radius R with the total charge Q spread uniformly over its perimeter. What is the potential difference between the point at the center of the ring and a point on its axis a distance 2R from the center?

R

Qk

R

Qk

R

Qk

RR

QkVVV

e

e

eeR

553.0

15

1

2 2202

P

R r

x

Q

2R

Electric Potential due to a charged conductor

(a) The excess charge on a conducting sphere of radius R is uniformly distributed on its surface.

(b) Electric potential versus distance r from the center of the charged conducting sphere.

(c) Electric field magnitude versus distance r from the center of the charged conducting sphere.

Pictures from Serway & Beichner

Example: a 1 C point charge is located at the origin and a -4 C point charge 4 meters along the +x axis. Calculate the electric potential at a point P, 3 meters along the +y axis.

q2q1

3 m

P

4 m

y

i 1 2P

i i 1 2

-6 -69

3

q q qV = k = k +

r r r

1×10 -4×10= 9×10 +

3 5

= - 4.2×10 V

Example: find the total potential energy of the system of three charges.

q2q1

3 m

P

4 mx

y

q31 2 1 3 2 3

12 13 23

q q q q q qU = k + +

r r r

-6 -6 -6 -6 -6 -6

91×10 -4×10 1×10 3×10 -4×10 3×10

U = 9 10 + +4 3 5

-2U = - 2.16 10 J

Example (Motion in Uniform Field)An insulating rod having linear charge density= 40.0 C/m and linear mass density = 0.100 kg/m is released from rest in a uniform electric field E = 100 V/m directed perpendicular to the rod as shown. (a) Determine the speed of the rod after it has traveled 2.00 m. (b) How does you answer to part (a) change if the electric field is not perpendicular to the rod ? (Ignore gravity)

Arbitrarily take V = 0 at the initial point, so at distance d downfield, where L is the rod length

V = Ed and Ue= LEd

(K +U)i =(K +U )f

0+0= 1

2Lv2 LEd

v =

2Ed

2(40.010 6C/ m)(100N / C)(2.00m)(0.100 kg/ m)

= 0.4 m/s

=

Potential Energy

Three point charges are fixed at the positions shown. The potential energy of this system of charges is given by

U = k

q1 q2

r12

q1 q3

r13

q2 q3

r23

+ +

Example: Electric Potential due to a dipole

An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a. The dipole is along the x-axis and is centered at the origin. (a) Calculate the electric potential at point P. (b) Calculate V and Ex at a point far from

the dipole. (c) Calculate V and Ex if P is

located anywhere between the two

charges.

V = qi

rik i

= k q

x- a x+ a

q=

2kqa

x2 - a2

(a)

(b)x >> a, V ~ 2kqa/x2 Ex =- dV/dx = 4kqa/x3

(c)V =

qi

rik i

= k q

a - x x+ a

q=

2kqx x2 - a2

Ex =dV

dx=

d

dx

2kqx

x2 - a2

= 2kq -x2-a2

x2 - a2 2

Electric Potential Energy: a point charge moves from i to f in an electric field, the change in electric potential energy is

WUUU if

Electric Potential Difference between two points i and f in an electric field: q

U

q

U

q

UVVV ifif

Equipotential surface: the points on it all have the same electric potential. No work is done while moving charge on it. The electric field is always directed perpendicularly to corresponding equipotential surfaces.

r

qrV

04

1)(

Potential due to point charges:

f

isdE

q

UV

0

Potential due to a collection of point charges:

n

i i

in

ii r

qVV

101 4

1

Potential of a charged conductor is constant everywhere inside the conductor and equal to its value to its value at the surface.

r

dqdVV

04

1

Calculatiing E from V: s

VEs

z

VEz

x

VEx

y

VEy

Electric potential energy of system of point charges:

r

qqVqU 21

02 4

1

summary

CapacitorCapacitora device that stores chargea device that stores charge

made of two conductors separated by an insulator

The amount of charge that a

capacitor can store depends on:

1. area of conducting surface

2. distance between the conductors

3. type of insulating material

link

Capacitancethe ratio of chargecharge to potential differencepotential difference

C = Q/VC = Q/V Q = CVQ = CV

The SI unit of capacitance

is the FaradFarad, FF, named in

honor of Michael Faraday Faraday.

One Farad of capacitance means that

one Coulomb of charge may be stored

in the capacitor for each Volt of

potential difference applied.

One Farad of capacitance means that

one Coulomb of charge may be stored

in the capacitor for each Volt of

potential difference applied.

Capacitor CircuitsCapacitor CircuitsSeriesSeries

1. reciprocal of the 1. reciprocal of the total capacitancetotal capacitance is the sum is the sum

of the reciprocals of the separate capacitorsof the reciprocals of the separate capacitors

1/C1/CTT = 1/C = 1/C11 + 1/C + 1/C22 + 1/C + 1/C33 + ... + ...2. 2. chargecharge is the same on each capacitor is the same on each capacitor

QQTT = Q = Q11 = Q = Q22 = Q = Q33 = ... = ...

3. 3. total potential differencetotal potential difference is the sum of each is the sum of each

VVTT = V = V11 + V + V22 + V + V33 + ... + ...In other words, in a series circuit,In other words, in a series circuit,

capacitance adds as reciprocals,capacitance adds as reciprocals,

charge stays the same,charge stays the same,

and voltage adds.and voltage adds.

CC11

CC22

CC33

EE = 12 V = 12 V

CCTT = 4.0 = 4.0 FF

VVTT = 12 V = 12 V

QQTT = 48 = 48 CC

CC11

CC22

CC33

C,C,

FFV,V,

VV

Q,Q,

CC

12

10

15

48

48

48

4.0

4.8

3.2

48124

1Ct

QV

ParallelParallel1. 1. total capacitancetotal capacitance is the sum of each is the sum of each

separate capacitor separate capacitor

CCTT = C = C11 + C + C22 + C + C33 + ... + ...2. 2. total chargetotal charge is the sum of the charges is the sum of the charges

on each separate capacitoron each separate capacitor

QQTT = Q = Q11 + Q + Q22 + Q + Q33 + ... + ...3. 3. potential differencepotential difference is the same across is the same across

each capacitoreach capacitor

VVTT = V = V11 = V = V22 = V = V33 = ... = ...In other words, in a parallel circuit, In other words, in a parallel circuit,

capacitance and charge add, capacitance and charge add,

but voltage stays the same.but voltage stays the same.

CC33

EE = 12 V = 12 V

CCTT = 22 = 22 FFVVTT = 12 V = 12 V

QQTT = 264 = 264 C C

CC11

CC22

CC33

C,C,

FFV,V,

VV

Q,Q,

CC

8

10

4CC22

CC11

12

12

12

96

120

48

•CCeqeqV V = = CC11VV+C+C22V+V+CC33VV

• 264264 = =8.128.12+10.12++10.12+4.124.12

Series Series CapacitorsCapacitors

VVtotaltotal= V= V11+V+V22+V+V33

VCVCeqeq= V= V11CC11= V= V22CC22= V= V33CC33 = Q = Q

1/C1/Ceqeq= 1/C= 1/C11+1/C+1/C22+1/C+1/C33

-- ++

+Q+Q

+Q+Q

+Q+Q

-Q-Q

-Q-Q

-Q-Q

Parallel Parallel CapacitorsCapacitors•QQtotaltotal= Q= Q11+Q+Q22+Q+Q33

•CCeqeqV = CV = C11V+CV+C22V+CV+C33VV

CCeqeq= C= C11+C+C22+C+C33

--

++

CC11 CC22 CC33

VV

QQtotaltotal QQ11 QQ22 QQ33

Q = CVQ = CV

Created by: azad sleman university of duhok 29

Capacitance

Capacitors in Parallel and Series:

Equivalent capacitance:

Example: (a) Find the equivalent capacitance of the combination as shown. Assume C1=12.0 F,

C2=5.30 F, C3=4.50 F. (b) A potential difference V=12.5 V is applied to the input terminals.

What is the charge on C1?

Solution: (a):

(b):Example: A 3.55 F capacitor C1 is charged to a potential difference V0=6.30 V, using a 6.30 V

battery. The battery is then removed and the capacitor is connected as in the figure to an uncharged 8.95 F capacitor C2. When switch S is closed, charge flows from C1 to C2 until the

capacitors have the same potential difference V. What is the common potential difference?

Solution:

(series)11

(parallel)11

n

i ieq

n

iieq CC

CC

312123

2112

111

CCC

CCC

F57.3

F3.17

123

12

C

C

V58.2C6.4412

123

12

1212123123

C

q

C

qVVCq

VCVCVC

qqq

2101

210

V79.121

10

CC

CVV

Capacitor & capacistancWe usually talk about capacitors in terms of parallel conducting plates

They in fact can be any two conducting objects

abV

QC

Units areVolt

Coulombfarad

1

11

The capacitance is defined to be the ratio of the amount of charge that is on the capacitor to the potential difference between the plates at this point

The magnitude of the electric field between the two plates is given by A

QE

00

We treat the field as being uniform allowing us to write A

dQEdVab

0

d

AC o

نه دنooاف ره بتيooدا ئooه م بفي دي قoooooooانوني جيكه ين

fffarad

d

AC o

0

35401054.31054.3

105/21085.8

39

312

coulombsVCqab

649

1054.3101054.3

coulombsVCqab

649

1054.3101054.3

mCAq /1077.12/1054.3/56

mvE /1021077.11036/659

0

mvdVEab

/102105/10/634

mCED /1077.11021085.85612

0

Cylindrical Capacitor: (Optional)The charge resides on the outer surface of the inner conductor and the inner wall of the outer conductor. Assume the length of the cylinder L>>b.

ab

LC

a

b

L

Qdr

rL

QEdrV

b

a

b

a

ln2

ln22

0

00

Spherical Capacitor: (Optional)

ab

abC

ba

Qdr

r

QEdrV

b

a

b

a

0

02

0

411

44

Example: An isolated conducting sphere whose radius R is 6.85 cm has a charge q=1.25 nC. (a) How much potential energy is stored in the electric field of this charged conductor? (b) What is the energy density at the surface of the sphere? (c) What is the radius R0 of an imaginary

spherical surface such that half of the stored potential energy lies within it?

Solution:

RRU

RR

qdrr

r

qdrruc

R

qEubn

R

q

C

qUa

R

R

R

R2

2

11

84

4

1

2

14:

J/m4.254

1

2

1

2

1:J103

82:

000

22

2

20

02

3

2

20

02

00

22

00

Potential Energy and Energy Density:

22

2

1

2CV

C

QU

202

1Eu

The electric potential energy of a capacitor is the energy stored in the electric field between the two plates (electrodes). It is the work required to charge the capacitor

The energy density is the potential energy per unit volume

1-In the figure, battery B supplies 12 V. (a) Find the charge on each capacitor first when only switch S1 is closed and (b) later when switch S2 is also closed. Take C1=1.0 F, C2=2.0 F, C3=3.0 F, and C4=4.0

F.

Home work

2-A parallel-plate capacitor of plate area A is filled with two dielectrics as shown in the figure. Show that the capacitance is

2210

d

AC

3-A parallel-plate capacitor of plate area A is filled with two dielectrics of the same thickness as shown in the figure. Show that the capacitance is

21

2102

d

AC

Calculating the Capacitance

Putting this all together, we have for the capacitance

d

A

V

QC

ab0

The capacitance is only dependent upon the geometry of the capacitor Example

A circuit consists of three unequal capacitors C1, C2, and C3

which are connected to a battery of emf The capacitors obtain charges Q1 Q2, Q3, and have voltages across their plates V1, V2,

and V3. Ceq is the equivalent capacitance of the circuit.

a) Q1= Q2 b) Q2= Q3 c) V2= V3

d) E = V1 e) V1 < V2 f) Ceq > C1

A detailed worksheet is available detailing the answers

The answer it is a green coulor

RedefinitionsWe now redefine several quantities using the dielectric constant

d

A

d

AKKCC 00

with the last two relationships holding for a parallel plate capacitor

Energy Density22

0 2

1

2

1EEKu

We define the permittivity of the dielectric as being

0 K

Capacitance:

Energy Stored in a CapacitorElectrical Potential energy is stored in a capacitorThe energy comes from the work that is done in

charging the capacitor

Let q and v be the intermediate charge and potential on the capacitor

The incremental work done in bringing an incremental charge, dq, to the capacitor is then given by

C

dqqdqvdW

ab

Force between two plate capacitor drFdW .

Energy density

ADF .2/2

AdxDdW

AxDW

.2/

.2/2

2

2/2

D

16.9 Energy Stored in a Capacitor

Average voltage during charging:Average voltage during charging:

Since Since VVfinalfinal is the applied voltage, we write is the applied voltage, we write VVaa==VV/2./2.

Energy stored (=work done by the battery): Energy stored (=work done by the battery): 

22finalinitialfinal

a

VVVV

ADpEdqEdfvqcqEW

CVqVqVW

.2/...2/12/

2

1

2

1

222

2a

0

The stored energy is: W = 0.5CV2 = 0.5 x .005 x1002 = 25 J

Example

d

A

- - - - -

+ + + +

Now suppose you pull the plates further apart so that the final separation is d1Which of the quantities Q, C, V, U, E change?

How do these quantities change?

Answers: Cd

dC

11 V

d

dV 1

1

Suppose the capacitor shown here is charged to Q and then the battery is disconnected

Ud

dU 1

1

Charge on the capacitor does not changeCapacitance DecreasesVoltage IncreasesPotential Energy Increases

Q:

C:

V:

U:

E: Electric Field does not change

Suppose the battery (V) is kept attached to the capacitor

Again pull the plates apart from d to d1Now which quantities, if any, change?

How much do these quantities change?

Answers:

Example

Cd

dC

11 E

d

dE

11 U

d

dU

11

Q:

C:

V:

U:

E:

Voltage on capacitor does not change

Capacitance Decreases

Charge Decreases

Potential Energy Decreases

Electric Field Decreases

Qd

dQ

11

Electric Field Energy DensityThe potential energy that is stored in the capacitor can be thought of as being stored in the electric field that is in the region between the two plates of the capacitor

The quantity that is of interest is in fact the energy density

dAuDensityEnergy

VC 22

1

where A and d are the area of the capacitor plates and their separation, respectively

Usingd

AC 0 and dEV we then have

202

1Eu

Even though we used the relationship for a parallel capacitor, this result holds for all capacitors regardless of configurationThis represents the energy density of the electric field in general

DielectricsMost capacitors have a nonconducting material between their platesThis nonconducting material, a dielectric, accomplishes three things1) Solves mechanical problem of keeping the

plates separated2) Increases the maximum potential difference allowed between the plates3) Increases the capacitance of a given capacitor

over what it would be without the dielectricSuppose we have a capacitor of value C0 that is charged to

a potential difference of V0 and then removed from the

charging sourceWe would then find that it has a charge of

00VCQ

We now insert the dielectric material into the capacitorWe find that the potential difference decreases by a factor K

K

VV 0

Or equivalently the capacitance has increased by a factor of K

0CKC

This constant K is known as the dielectric constant and is dependent upon the material used and is a number greater than 1

16.10 Capacitors with Dielectrics• A dielectric is an insulating material that, when placed

between the plates of a capacitor, increases the capacitance– Dielectrics include rubber, plastic, or waxed paper

• C = κCo = κεo(A/d)– The capacitance is multiplied by the factor κ when the

dielectric completely fills the region between the plates

Capacitance in presence of a dielectric:

d

AC

CV

Q

/V

Q

V

QC

0

0

0000

Since >1, the dielectric enhances the capacitance of the capacitor!

(a)Electric field lines inside an empty capacitor

(b)The electric field produces polarization

(c) The resulting positive and negative surface charges on the dielectric reduce the electric field within the dielectric

E0

E=E0/or

V=V0/

+Q0-Q0

Reasoning:

Dielectric constant

Capacitors with Dielectrics

The value of depends on the nature of the dielectric material, as the table below indicates

Capacitors Designs

(a)Paper capacitor(b)High-voltage oil

capacitor(c)Electrolytic capacitor

Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is

filled with a dielectric.

Compare the voltages of the two capacitors.

a) V1 > V2 b) V1 = V2 c) V1 <

V2

Example

We have that Q1 = Q2 and that C2 = KC1

We also have that C = Q/V or V= Q/C

Then1

11 C

QV and 1

1

1

2

22

1V

KKC

Q

C

QV

Electric Current

The electric current is the amount of charge per unit time that passes through a plane that pass completely through the conductor.

The SI unit for current is a coulomb per second (C/s), called as an ampere (A)

Avnqt

QI

d

nqAIv

a/

Microscopic Description of Current: Math

m

EnqJ

2

•Assume uniform motion and density of charge carriers

A

•The charge in a wire of length L can be calculated

L

q=(nAL)ev, for electrons

•The total charge moves through a cross-section in:

t=L/v ; v is the drift velocity

A

•Assume uniform motion and density of charge carriers

•I=q/t=nALev/L =nAev

•This implies (J=I/A) that J=(ne)v

ne is the charge carrier density

Conductivity•In most materials, electric field is required to make the current move

•the current is proportional to the electric field, and the conductivity

•The conductivity is a property of the material

EJ Empirically,

EJ

•the resistivity is the reciprocal of the conductivity, nothing more!

Current I

Electric Field E

Adobe Acrobat Document

Direction of current

• A current arrow is drawn in the direction in which positive charge carriers would move, even if the actual charge carriers are negative and move in the opposite direction.

• The direction of conventional current is always from a point of higher potential toward a point of lower potential—that is, from the positive toward the negative terminal.

A complete circuit is one where current can flow all the way around. Note that the schematic drawing doesn’t look much like the physical circuit!

Electric current

current density J If the current is constant and perpendicular to a surface, then and we can write an expression for the magnitude of the current density

J i

AJ

i

A nevJ

eqnqvJ

Daft speed

Current DensityWhen a conductor is in electrostatic equilibrium, its conduction electrons move randomly with no net direction. In the presence of electric field, these electrons still move randomly, but now tend to drift with a drift speed, in the direction opposite to that of the applied electric field that causes the current. The drift speed is tiny compared to the speed in the random direction. For copper conductors, typical drift speeds are from 10-5 to 10-4 m/s as compare to 106m/s for random motion speedsFor convenience, the drift of positive charged particles is in the direction of the applied electric field E. Assume all particles move at the same drift speed and the current density is uniform across the wire’s cross sectional A. The number of particles in a length L of the wire is nAL, where n is the number of particles per unit volume. Therefore the total charged particles in the length L, each with charge e, is

q=(nAL)e

Since the particles all move along the wire with speed vd this total charge q

moves though any cross-section of the wire in the time interval

t = L / vd

In vector form

Suppose, instead that the current density through a cross- section of the wire varies with radial distance r as J = ar2 which a=3 x 1011A/m2 and r is in meters. In this case what is the current through the outer portion of the wire between radial distance R/2 and R ?Since the current density is NOT

uniform,

We have to resort to the integral relationship

I = J • dA and integrate the current density over the portion of the wire from r = R/2 to R

The current density vector J (along the wires length) and the differential area vector dA (perpendicular to the cross section of the wire) have the same direction. Therefore We need to replace the differential area dA with something we can integrate between the limits r=R/2 and r=R. The simplest replacement is the area 2rdr of a thin ring of circumference 2r and thickness dr.

We can then integrate with r as the variable

Problem

Problem

From the previous slide, we must solve the integral

67

Example - current through a wire• The current density in a cylindrical wire of radius R=2.0

mm is uniform across a cross section of the wire and has the value 2.0 105 A/m2. What is the current i through the outer portion of the wire between radial distances R/2 and R?

• J = current per unit area = di / dA

RArea A’ (outer portion)

Current through A’

The resistance

• The resistance (R) is defined as the ratio of the voltage V applied across a piece of material to the current I

through the material: R=V/i..SI Unit of Resistance: volt/ampere

(V/A)=ohm(Ω)

69

Resistance vs Resistivity• The conducting properties of a material are

characterized in terms of its resistivity.• We define the resistivity, , of a material by the ratio

E: magnitude of the applied field

J: magnitude of the current density

E

J

V

m

A

m2

=Vm

Am

R L

AR

L

A

For a wire

is measured in m

L – length of the conductor

A – its area.

• The units of resistivity are

resistivity = resistance x cross section area/ length:

Example III–2. Problem 17.17 (Page 590) from the Serway

& Voile textbook: A wire 50.0 m long and 2.00 mm in diameter

is connected to a source with a potential difference of 9.11 V,

and the current is found to be 36.0 A. Assume a temperature of

20C and, using Table 17.1, identify the metal of the wire.

Example about Resistivity

253.0

36

12.9

A

V

I

vR

m

m

m

L

dR

L

RA

.1059.1

)50(4

)102()253.0()4/(

8

232

It is a silver

RESISTIVITY

25 x 103 Ω

±10%

RESESTANE AND RESISTOR

بفي ريكي تooو دشooي هزمooارا مقooاومي بكooه ي بي بكارهينانا جهازي بجه ر بينه

73

Resistance and Temperature• When electrons move through the conductor

they collide with atoms:– Resistivity grows with temperature ( more

collisions)

– resistivity measured at some reference temperature T0

– temperature coefficient of resistivityWhen electrons move through the conductor they collide with atoms:

Temperature of the conductor increases because of the current (through collisions)

Electrical energy is transformed into thermal energy

Resistors dissipate energy

Power – energy per unit of time- (in W=J/s) dissipated by a resistor

s the resistivity at some temperature T.

)](1[00

TTaRR

If a wire is of constant cross-sectional area A and length L with

respect to change in temperature, we can write

Example III–3. At 40.0C, the resistance of a segment of gold

wire is 100.0 . When the wire is placed in a liquid bath, the resistance

decreases to 97.0 . What is the temperature of the bath?

(Hint: First determine the resistance of the gold wire at room temperature.)

Power Consumed by a Resistor

E–

+

V = 0

V = E

I = E/RR

dUP

dt d

Q Vdt

dQV

dt I V

P I V

22 VP RI

R

,WATT,W

Solve the current, v-drop across each resistor, and Req for the circuit.

12v 12v

Example: A rectangular block of iron has dimension 1.2 cm1.2 cm15 cm. (a) What is the resistance of the block measured between the two square ends? (b) What is the resistance between two opposite rectangular faces?

Solution:

Power:

Example: A wire has a resistance R of 72 . At what rate is energy dissipated in each of the following situations? (1) A potential difference of 120 V is applied across the full length of the wire. (2) The wire is is cut in half, and a potential difference of 120 V is applied across the length of each half.

Solution:

Example: A wire of length L=2.35 m and diameter d=1.63 mm carries a current i of 1.24 A. The wire dissipates electrical energy at the rate P=48.5 mW. Of what is the wire made?

Solution:

78

Clicker QuestionYou have three cylindrical copper conductors. Rank

them according to the current through them, the greatest first, when the same potential difference V is

placed across their lengths .A: a, b, c

B: a and c tie, then b

C: b, a, c

D: a and b tie, then c

B: a and c tie, then b

OHMS LAW

Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

Ohm’s Law

The ratio of voltage to current is called the resistance: or

Unit of resistance: the ohm, Ω. 1 Ω = 1 V/A.

Microscopic View of Ohm’s LawThe motion of the electrons in an electric field E is a combination of motion due to random collisions and that due to E. The electron random motions average to zero and make no contribution to the drift speed. So the drift speed is due only to the electric field.

In the average mean free time between collisions, the average electron will acquire a speed of

Recalling that

If an electron of mass m is placed in electron field, it will undergo an acceleration.

, we can write the above result as

Solve for E and we have

Recalling that The above equation becomes

Microscopic Description of Ohm’s law

•Microscopically current is due to the movement of charge carriers

When a field is applied, the symmetry of the “motion” of the electrons is

broken and there is a net drift.

m

EnqJ

2

nqvJ

We can rewrite this in a different form,

atv m

Fa qEF

Put this all together and,

We also now know what the conductivity and

resistivity are.

82

• Coefficient in this dependence is called resistance R

• Resistance is measured in Ohm ( = V/A)

Ohm’s law• Electric current is proportional to voltage.

VI IRV

R

V

I

Through in technical