PH 222-2C Fall 2012

Electric Potential

Lectures 5-6

Chapter 24(Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 24Electric Potential

In this chapter we will define the electric potential ( symbol V )associated with the electric force and accomplish the following tasks:

Calculate V if we know the corresponding electric field. Calculate the electric field if we know the corresponding potential V.Determine the potential V generated by a point charge. Determine the potential V generated by a discrete charge distribution. Determine the potential V generated by a continuous charge distribution. Determine the electric potential energy U of a system of charges. Define the notion of an equipotential surface. Explore the geometric relationship between equipotential surfaces and electric field lines. Explore the potential of a charged isolated conductor.

2

A

B

O

x. . .xi x xf

F(x)In Chapter 8 we defined the

associated with a conservative force as the negative value of the work that the force must do on a particle to take

UW

Electric Potential Energy change in potential

energy

0

it from an initialposition to a final position .

( )

Consider an electric charge moving from an initial position at point to a finalposition at point under the inf

f

i

i f

x

f ix

x x

U U U W F x dx

qA

B

0

0

luence of a

known electric field . The force exerted

on the charge is .

f f

i i

E

F q E

U F ds q E ds

( )f

i

x

x

U F x dx

0

f

i

U q E ds

3

A

B

0

0

The change in potential energy of a charge

moving under the influence of from point A

to point is: .

Please note that depends on the val

f

f ii

q

E

B U U U W q E ds

U

The Electric Potential

V

0ue of .q

00 0

We define the in such a manner so that it is independent

of : Here .

In all physical problems only changes in are involved. Thus w

f

f i f ii

U Wq V V V V V V E dsq q

V

electric potential

V

e can definearbitrarily the value of at a reference point, which we choose to be at infinity:

0. We take the initial position as the generic point with potential :

The pote.

f

P

P

P

VV V P

V s

V

E d

ntial depends only on the coordinates of and on .

PV P E

P

PV E ds

4

O

0

Definition of voltage :

Units of :

Consider a point charge placed at the origin. We

J/C

wi

, known as the volt

ll usethe defini

W

q

Vq

V

Potential Due to a Point Ch

SI Units of :

a e

rg

V

P

20

20

tion given on the previous page to determine the potential at point a distance from .

cos 0

The electric field generated by is:

4

4

R

PR R

P

V P R O

V E ds Edr Edr

qqE

r

q drVr

2

0 0

1 14

1

4

R

PR

dr

q qVr

x x

R

0

14P

qVR

5

6

Example. The potential of a point charge for a zero reference potential at infinity

A point charge of q=4.0x10-8C creates a potential at a spot 1.2 m away.

The potential is (a) V=kq/r=

=(8.99x109)(+4.0x10-8)/1.2=+300V when the charge is positive

and (b) -300V when the charge is negative

7

The acceleration of positive and negative charges

A positive charge accelerates from a region of higher electric potential toward a region of lower electric potential

A negative charge accelerates from a region of lower electric potential toward a region of higher electric potential

8

Electrostatic Potential of a point charge

9

The electric potential energy of a point charge q for non-uniform electric field

10

One electron volt is the magnitude of the amount by which the potential energy of an electron changes when the electron moves through the potential difference of 1 volt

11

12

A headlight connected to a 12-V battery

13

-2.3x10-18 J

-6.5x10-19 J

done by the electric force

Work to be done against the electric force = +10 eV

14

15

r1r2

r3

P

q1

q3

q2Consider the group of three point charges shown in thefigure. The potential generated by this group at anypoint is calculated using the principle of super

VP

Potential Due to a Group of Point Charges

1 2 3

31 21 2 3

0 1

1 2

0 1 0 2 0

0 2 0 3

1 2 3

position. We determine the potentials , and generated

by each charge at point : 1 1 1

1

, , 4 4 4

We add the three ter

1 14 4 4

ms:

V , V VP

qq qV V Vr r r

V V V Vqq qV

r r

1.

2.

3

3r

1 2

10 1 0 2 0 0

The previous equation can be generalized for charges as 1 1 1 1...

4 4 4

follow

4

s:n

n i

n i

q qq qVr r r

n

r

1 2 3V V V V

16

A

B

C

p

Consider the electric dipole shown in the figure.We will determine the electric potential created at point by the two charges of the dipole using superpo

V P

Example : Potential Due to an Electric Dipole

( ) ( )( ) ( )

0 ( ) ( ) 0 ( ) ( )

sition.Point is at a distance from the center of the dipole.Line makes an angle with the dipole axis:

1 .4 4

We assume that , where

P r OOP

r rq q qV V Vr r r r

r d

( ) ( )

2( ) ( ) 2 2

0 0

is the charge separation.From triangle we have: cos .

cos 1 cosAlso: ,4 4

where the electric dipole moment.

dABC r r d

q d pr r r Vr r

p qd

20

1 cos4

pVr

17

.Pr

dq

Consider the charge distribution shown in the figure.In order to determine the electric potential createdby the distribution at point we use the pr

VP

Potential Due to a Continuous Charge Distribution

inciple ofsuperposition as follows:

We divide the distribution into elements of charge .For a volume charge distribution, .For a surface charge distribution, .For a linear charge distr

dqdq dVdq dA

1.

ibution, .dq d

0

0

1 We determine the potential created by at : .4

1 We sum all the contributions in the form of the integral: .4

The integral is taken over the whole charge dist

dqdV dq P dVr

dqVr

2.

3.

Note 1 : ribution. The integral involves only scalar quantities. Note 2 :

0

14

dqVr

18

dqO A

Potential created by a line of charge of length and uniform linear charge density λ at point . Consider the charge element at point , adistance from . From triangle we

L Pdq dx A

x O OAP

Example :

2 2

2 2

2 2

2 20 0

2 20 0

2 2

00

2 2

0

have:

. Here is the distance .The potential created by at is:

1 14 4

4

ln4

l ln4

ln

n

L

L

r d x d OPdV dq P

dq dxdVr d x

dxV

dx x d xd x

d x

V x d x

V L L d d

19

( )F

( )F

2

2 2

Many molecules such as H O have a permanent electricdipole moment. These are known as "polar" molecules. For others, such as O , N , etc. the electric dipole moment is zero. The

Induced Dipole Moment

se are known as "nonpolar" molecules.One such molecule is shown in fig. . The electric dipolemoment is zero because the center of the positivecharge coincides with the center of the negative char

ap

ge.

In fig. we show what happens when an electric field is applied to a nonpolar molecule. The electric forces on the positive and negative charges are equal in magnitudebut opposite in direction.

b E

As a result the centers of the positive and negative charges move in oppositedirections and do not coincide. Thus a nonzero electric dipole moment appears. This is known as "induced" electric dipol

p

e moment, and the moleculeis said to be "polarized." When the electric field is removed disappears.p 20

A collection of points that have the samepotential is known as an equipotential surface. Four such surfaces are shown in

the figure. The work done by as it movesa charge be

Eq

Equipotential Surfaces

tween two points that have apotential difference is given by

. V

W q V

I

II

III 1 2

IV 1 2

For path I : 0 because 0.For path II: 0 because 0.For path III: .

For path IV: .

When a charge is moved on an equipotential surface 0the wo

W VW VW q V q V V

W q V q V V

V

Note :rk done by the electric field is zero: 0.W

W q V

21

SA B

V

q

F

E

r

Consider the equipotential surface at potential . A charge is moved

by an electric field from point to point along a path

V q

E AB

The Electric Field is Perpendicular to the Equipotential Surfaces

E

. Points and and the path lie on .

rA B S

Let's assume that the electric field forms an angle with the path .

The work done by the electric field is: cos cos .We also know that 0. Thus: cos 0, where

0, 0

E r

W F r F r qE rW qE r

q E

, 0. Thus: cos 0 The correct picture is shown in the fi

90 .gure below.

r

SV

E

22

Examples of Equipotential Surfaces and the Corresponding Electric Field Lines

Uniform electric field Isolated point charge Electric dipole

0 0

. Assume that is constant constant.4 4

Thus the equipotential surfaces are spheres with their center at the point charge

and radius 4

q qV V rr V

qr

Equipotential surfaces for a point charge :q

0

.V

23

A

B

V V+dV

Now we will tackle the reverse problem, i.e., determine if we know . Consider two equipotential surfaces that correspond to the values and

E V

E VV

Calculating the Electric Field from the Potential

0

separated by a distance as shown in the figure. Consider an arbitrary directionrepresented by the vector . We will allow the electric field to move a charge from the equipotential sur

V dVds

dsq

face to the surface .V V dV

0 0

0 0

The work done by the electric field is given by: (eq. 1). Also cos cos (eq. 2)

If we compare these two equations we have:

cos cos .

From triangle we see th

W q dV W Fds Eq ds

dVEq ds q dV Eds

PAB

at cos is the component of

along the direction . Thus: . We have switched to the

partial derivative symbols to emphasize that

involves only the variation of along a spe

s

s

s

E E EVs Es

VEs

V

cific axis.s sVEs

24

A

B

V V+dV

We have proved that .sVEs

s

VEs

The component of in any direction is the negative of the rate at which the electric potential changes with distance in this direction.

E

If we take to be the - , -, and -axes we get:

If we know the function ( , , )

we can determine the components of

and thus the vector itself :

; ; x y z

s x y z

V x y z

E

V V VE E Ex y

E

z

ˆˆ ˆV V VE i j kx y z

25

For the simple situation in which the electric field is

uniform , where is perpendicular to the

equipotential surface.

V ss

E

E

We define as the work required to assemble thesystem of charges one by one, bringing each chargefrom infinity to its final position.Using the above def

UPotential Energy of a System of Point ChargesU

inition we will prove that for a system of three point charges is given by:U

2 3 1 31 2

0 12 0 23 0 134 4 4q q q qq qU

r r r x

y

O

q1

q2

q3

r12

r23

r13

, 10

Each pair of charges is counted only once.For a system of point charges the potential en

1 .

ergy is given by:

Here is the separation between and 4

ni j

i j iji

i i

j

i

j

q qn q U

r q qUr

Note :

.

The summation condition is imposed so that, as in the case of three point charges, each pair of charges is counted only once.

j

i j

26

1

1

2

2 2

1 1 22

0 12 0 12

3

3 3

1 2

0 13 23

1 33

0

Bring in : 0

(no other charges around)Bring in : (2)

(2)4 4

Bring in : (3)

1(3

)4

14

qW

qW q V

q q qV Wr r

qW q V

q qVr r

q qWr

Step 2 :

Step 3

Step :

:

1

2 3 1 31 2

2 3

13 23

1 2 3

0 12 0 23 0 134 4 4q q q qq qW

r r

q qr

W W W W

r

O

x

y

q1

Step 1

x

y

O

r12q1

q2

Step 2

Step 3

x

y

O

r12

r13

r23

q1

q2

q3

27

28

29

Energy of a system of charged conductors

30

31

=1/40(q2/L+q3/2L +q4/L)

path

A

B

conductor

0E

We shall prove that all the points on a conductor(either on the surface or inside) have the same potential.

Potential of an Isolated Conductor

Consider two points and on or inside a conductor. The potential difference between these two points is given by the equation:

.

We already know that the electrostatic field

B AB

B AA

A BV V

V V E d S

inside a conductor is zero.Thus the integral above vanishes, and for any two pointson or inside the conductor.

B A

EV V

A conductor is an equipotential surface.

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We already know that the surface of a conductor is an equipotential surface. We also know that the electric field lines are perpendicular to theequipote

Isolated Conductor in an External Electric Field

ntial surfaces.

From these two statements it follows that the electric field vector is perpendicular to the conductor surface, as shown in the figure. All the charges of the conductor reside on the surface and a

E

in

out0

rrange themselves in such a way so that the net electric field inside the conductor

The electric field just outside the conductor

0.

is .:

E

E

33

Electric Field and Potential in and around a Charged Conductor : A Summary

in

out0

All the charges reside on the conductor surface. The electric field inside the conductor is zero: 0.

The electric field just outside the conductor is: .

The electric fi

E

E

1. 2.

3.

4. eld just outside the conductor is perpendicular to the conductor surface.

All the points on the surface and inside the conductor have the same potential. The conductor is an equipotential 5.

surface.

in 0E

out0

ˆE n

E

E

n̂

n̂

34

R

0

0

1For , .4

1For , .4

qr R VRqr R Vr

20

For , 0.1For , .

4

r R Eqr R Er

Electric Field and Electric Potentialfor a Spherical Conductor of Radius and Charge

Rq

Outside the spherical conductor, the electric fieldand the electric potential are identical to that of a point charge equal to the net conductor charge and placedat the center of the sp e e.

h r

Note :

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