Chapter 3

ELECTRIC POTENTIAL

Recommended Problems:

5,7,9,17,19,21,23,31,33,43,45,46,47,49,58,66,68,69,71.

In this chapter we are going to discuss the following topics:

Electric Potential

Electric Potential for Point Charges

Electric Potential for Continuous Charge Distributions

Electric Potential Energy

Electric Filed Lines

Potential Difference and Electric Potential

Consider a charged particle of charge q in a region of an electric

field E. This field exerts an electric force on the particle given by

F=qE.

When the particle moves from initial point i to final point f this

force does work on it according to

Since the electric force is conservative, we can associate a

potential energy with this force according to

f

i

dqW lE

WU

f

iif dqUU lE

Let us now define the electric potential V as the electric potential

energy per unit charge, i.e.,

q

UV

The electric potential difference between the points i and f is then

f

iif dVV lE

The potential difference Vf - Vi is defined as the work required to

move a positive unit charge from point i to point f.

Since it is only the change in potential between two points that

has physical sense, it is often convenient to choose a reference

point for zero potential.

This reference point is usually chosen to be at infinity, i.e., V =0

Letting the point i to be and designate point f as point p, the

electric potential at any point p becomes

p

p dV lE

The electric potential at any point p can be defined as the work

required to bring a positive unit charge from infinity to that point. In

this sense the potential at a point is the potential difference

between that point and a point at infinity.

It follows that the electric field has a unit of volt per meter (V/m)

with 1 N/C = 1 V/m

The SI unit of V is joules per coulomb (J/C), usually represented

by a special unit called volt (V), i.e., 1 V = 1 J/C

From the above definitions, we conclude that

The electric potential is a scalar quantity.

Example 25.2 A proton is released

from rest in a uniform e.f. that has a

magnitude of 8.0 × 104 V/m. The proton

undergoes a displacement of 0.5 m in the

direction of E.

Solution

a) Find VB - VA

b) Find UB - UA

c) Find the speed of the proton at point B.

B

AAB dVVV lE

B

AAB dlEVV

Since E and dl are parallel

E

A B 0.5 m

Since E is uniform it can be taken out from the integral sign

EddlEVB

A

V44 100.45.0100.8

From the minus sign we conclude that VA VB.

VqU b)

J15419 104.6100.4106.1

c) 0UK

0104.60 15221 mv

m/s108.2

1067.1

104.62 6

27

15

v

• Test Your Understanding (1)

In the figure, two points A and B are

located within a region in which there

is an electric field. The potential

difference V =VB - VA is:

a) zero b) positive c) negative d) can't be determined

Referring to the same figure, if an electron is transferred from

point A to point B the change in potential energy U = UB - UA is:

a) zero b) positive c) negative d) can't be determined

ELECTRIC POTENTIAL DUE TO POINT

CHARGES

Consider an isolated

point charge q. We

want to find the electric

potential due to this

charge at a point. We

have

q

A B rA

dl

E

rB

r

B

AAB dVV lE

The electric field due to the charge q at a distance r is

r̂2r

qkE

It is clear that E and dl are in the same direction, so

2r

drKqEdldlE

AB

B

AAB

r

kq

r

kq

r

drkqVV

2

The first term of the right hand side represents the potential at

point B (VB) and the second term represents the potential at point

A (VA).

The electric potential at a point a distance r from a point charge q

is then obtained if rA , that is,

r

qkV

As it is clear from the above equation, V is positive for positive q

and negative for negative q.

The electric potential due to a group of point charges at a point is

the algebraic sum of the electric potentials due to each charge

individually.

That is, for a group of N point charges we have

N

i i

i

r

qkV

where ri is the distance from the ith charge qi to the point in

question.

Do not forget that the sum is algebraic sum and not vector sum

like that used to calculate the electric filed due to a group of point

charges. This fact gives an important advantage of potential over

electric field.

Example 25.3 Two point charges of

q1= 2.0 C, q2= -6.0 C, are arranged as

shown. Find

a) the electric potential at the point p.

b) the change in potential energy of the

system plus a charge q3= 3.0 C that

moves from to point P.

q2= - 6.0 C

q1=2.0 C

p 4.0 m

5.0 m 3.0 m

Solution

2

2

1

1

r

q

r

qkV V1029.610

5

6

4

2109 369

PVqWU 3

b) The change in P.E is equal to the work required, by an external

agent, to bring q3 from to point P, i.e.,

J1089.11029.6100.3 236

POTENTIAL ENERGY OF A SYSTEM OF

POINT CHARGES

Let us calculate the work required to assemble a system of three point charges.

To do so we have to calculate the work required to bring, from

infinity, each charge one by one.

As the potential at a point is equal to the work required to bring a

positive unit charge from infinity to that point. Therefore, the work

required to bring a charge q from infinity to a point must equal to

the potential at that point multiplied by q, i.e.,

pp qVW

Now, no work is required to place the

first charge q1 at a given position

because there is no electric potential at that position W1=0 .

q1

q2 r12

q3

r23

r13

Next we place a second charge q2 at a position r12 from q1.

This requires a work q2V1 where V1 is the potential at the location of q2 due to q1, or

Next we place q3 at a position r31 from q1 and r32 from q2.

12

212

r

qqkW

We now must do work given as q3V12, where V12 is the potential at

the location of q3 due both q1 and q2, i.e.,

32

2

31

133

r

q

r

qkqW

The total work required to assemble a system of three charges is

then

321 WWWW

This work is stored as an electrostatic energy in the system, so we

write

23

32

13

31

12

21

r

qq

r

qq

r

qqkWU

The potential energy of a system of N charges can be calculated

in a similar fashion. For 4-charges, for example we have

34

43

24

42

23

32

14

41

13

31

12

21

r

qq

r

qq

r

qq

r

qq

r

qq

r

qqkWU

For a simple system of only two point charges the potential energy

is given by

12

21

r

qqkU

Note that if the charges are of the same sign U is positive which

means that the force is repulsive, as expected.

If the charges are of the opposite sign U is negativetive which

means that the force is attractive, which is the case.

Example 25.4 Find the potential energy of

the system given in the previous example.

q2= - 6.0 C

q1=2.0 C

4.0 m

0.5 m 3.0 m

Solution

q3=3.0 C

23

32

13

31

12

21

r

qq

r

qq

r

qqkWU

5

10)0.6)(0.3(

0.4

10)0.3)(0.2(

0.3

10)0.6)(0.2(109

1212129

J10489.5 2

E. POTENTIAL DUE TO CONTINUOUS

CHARGE DISTRIBUTION

If a charge is continuously distributed in a given region, two

methods will be followed:

1- If the charge distribution is well known we can proceed as we

do for calculating the electric field discussed:

choosing an element of charge dq,

finding the potential dV due to this element,

and then integrating over the entire charge distribution.

2- If the electric field of the charge distribution can be calculated

easily (Using Gauss’ law for instant), we can use the Equation

p

p dV lE

Example 25.7 A wire of length L

has a charge Q uniformly distributed.

Find the electric potential at a point

along its axis and a distance a from

one end.

Solution

We divide the wire into small elements each of length dx and charge dq.

Since each element can be considered as a point charge, the

electric potential dV due to one of these elements a distance r from p is

r

dqkdV

22 ax

dqk

L a

dq dx

P

r

x

Since we are dealing with a scalar

quantity we can integrate the above

expression directly to get the

potential due to the whole wire. The

result is

22 ax

dqkV

Using the relation between dq and dx, i.e., dxdq

L

ax

dxkV

022

a

aLLkV

22

ln

a

aLL

L

Qk22

ln

L a

dq dx

P

r

x

22 ax

dqkV We note that x varies from 0 to L

Example 25.5 A ring of radius R has

a uniform charge distribution of magnitude

Q. Calculate the electric potential along the

axis of the ring a distance D from the

center of the ring.

Solution

Again we divide the wire into small elements each of charge dq.

Each element can be considered as a point charge. So the electric potential dV due to one of these elements at P is

r

dqkdV

22 DR

dqk

Integrating with all the constants are

taken out of the integrand we get

R

x P D

r

dq

Q

dqDR

kV

022 22 DR

kQ

At the center of the ring (D=0)

the electric potential is given by R

QkV

Example 25.6 A disk of radius R

has a uniform charge density .

Calculate the electric potential along the

axis of the disk at a point p lying a

distance D from the center of the disk.

Solution

Here we divide the disk into thin rings each of radius r and thickness dr.

From the result of the previous example, the potential due to one

element with radius r is

22 Dr

dqkdV

22 Dr

dqkV

Using the relation between dq and r, i.e.,

R

P D r

dr

dAdq rdr 2

R

Dr

rdrkV

0 22 21

2 RDrk 0

222 DDRk 222

Example 25.8 An insulating solid sphere of

radius R has a charge Q uniformly distributed within

its volume. Find the electric potential at a point

a) outside the sphere r R,

b) b) inside the sphere r R.

r

Q

R

Solution

a) For outside the sphere r R, we have

r

out dV lE

But E here is the e.field outside the sphere given by

Rrr

Qout rE ˆ

4 2o

r

outr

dQV

2o

ˆ

4

rl

r

outr

drQV

2o4

r

rkQ

1

r

Qk

Note that this result is identical to that of a point charge. This

means that the problem can be treated as the entire charge was

concentrated at the center of the sphere.

b) For inside the sphere r R, we have

r

in dV lE

But in this case the range from to

the point in question enclose two

regions with different electric fields:

The outside region (r R) which extend from to R,

and the inside region (r R) which extend from R to r.

Therefore, we have to divide the integration into two parts as

r

Rin

R

in ddV lElEout

The e.field outside and inside the sphere given by

Rrr

Qout rE ˆ

4 2o

RrR

Qrin rE ˆ

4 3o

Q

R r

r

R

R

in rdrR

Q

r

drQV

3o

2o 44

2

2

o

38 R

r

R

QVin

Performing the integrals and rearrange we get

Note that the two expressions of V1 and V2 give the same result

for r=R (at the surface) which is

R

QV

o4

ELECTRIC POTENTIAL AND CONDUCTORS

In the previous chapter we have mentioned some properties of a

conductor in electrostatic equilibrium with the help of Gauss’ law.

Some of these properties are:

(i) The electric field inside the conductor is zero.

(ii) The electric field on the surface of the conductor is

perpendicular to the surface. Now

f

iif dVV lE

If the initial and the final points lie inside a conductor and using

the first property we conclude that the electric potential is constant

inside the conductor.

Furthermore, since E is always normal to dl on the surface, the

electric potential is also constant on the surface of the conductor.

Another important property of a conductor thus follow

The electric potential inside any conductor in equilibrium is

constant and equal to the potential on its surface.

To prove this property explicitly, let us consider a conducting

sphere of radius R and charge Q. The electric potential at a point

inside the sphere is

r

R

Rr

dddV lElElE inout=

But Ein=0 and rE ˆ4 2

o

outr

Q

R

QV

o4

which is the value of the potential at the surface of the conductor.

R

R

r

V

The figure shows the variation of V as

a function of the radial distance r for a

conducting sphere of radius R.

If two or more conducting objects are

connected by a conducting wire, the

conductors are no longer separate but

can be considered as a single conductor.

This means that the electric charges will transfer from the

conductor of higher potential to that of lower potential until the

equilibrium condition is achieved.

Therefore, if two or more conductors are connected and

equilibrium is achieved, they must be at the same electric

potential.

In analogue with the electric filed lines, the electric potential can

be represented by equipotential surfaces.

A surface with all its points are at the same electric potential

is called equipotential surface.

From this definition, it follows that no work is done by the electric

field in moving a charged particle between two points on the same

equipotential surface.

This means that the electric filed lines must be perpendicular to

the equipotential surfaces.

The surface of any conductor forms an equipotential surface.

• Test Your Understanding (2)

A conducting sphere is charged with Q. The work required to

transfer a charge q from the center of the sphere to a point on the

surface is:

a) zero b) positive c) negative d) can't be determined

A system of two point charges q1 and q2. If the potential energy of

the system is negative then the force between the two charges is:

a) attractive b) repulsive c) zero d) can't be known

Example 25.9 Two conducting charged spheres with radii R1

and R2 are separated by a distance much larger than the radius of

either sphere. The two spheres are connected by a conducting

wire. The charges on the spheres in equilibrium are q1 and q2.

Solution

Find the ratio of the magnitudes of the e. fields at the surfaces of

the spheres, and the ratio of the charge densities on the surfaces

of the spheres.

Since the spheres are connected by a conducting wire, the

potential is the same for both spheres, i.e.,

21 VV

2

2

1

1

R

qK

R

qK

R1 R2

2

2

1

1

R

q

R

q

The charges are distributed over the surface of the spheres, so

we have

21

11

4 R

q

22

22

4and

R

q

212

221

2

1

rq

rq

21 qqSubstituting for the ratio

1

2

2

1

r

r

It is clear from this result that the charge density is greatest on the

small sphere as expected.

Now, since the e. field on the surface of a conductor is equal to

o

2

1

2

1

E

E

1

2

r

r

That is, the field is more intense near the smaller sphere.

• Test Your Understanding (3)

Two concentric spherical shells of radii a

and b are charged to a total charge Q.

The electric potential on the inner shell is

a) b) c) d) zero

Referring to the previous configuration, which of the following

statements is wrong:

a) q1=0 b) q2= Q

c) The two shells have the

same potential.

d) The outer shell has the larger

potential.

aQK bQK aqK 1

• Test Your Understanding (4)