Chapter 24
Electric 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.
(24-1)
xf
U    F ( x)dx
Electric Potential Energy
In Chapter 8 we defined the change in potential
xi
F(x)
.O
.
xi
.
energy U associated with a conservative force as the
negative value of the work W that the force
x
xf
x
must do on a particle to take it from an initial
position xi to a final position x f .
xf
U  U f  U i  W    F ( x)dx
f
U  q0  E  ds
xi
i
Consider an electric charge q0 moving
from an initial position at point A to a final
position at point B under the influence of a
A
known electric field E. The force exerted
on the charge is F  q0 E.
B
f
f
i
i
U    F  ds  q0  E  ds
(24-2)
Checkpoint 1:
In the figure, a proton moves from point i to point f in a uniform electric
field directed as shown.
a) Does the electric field do positive or negative work on the proton?
b) Does the electric potential energy of the proton increase or decrease?
c) Does the force do positive or negative work?
d) Does the proton move to a point of higher or lower potential?
P
VP    E  ds
The Electric Potential V

The change in potential energy of a charge q0
A
moving under the influence of E from point A
f
to point B is: U  U f  U i  W  q0  E  ds .
i
B
Please note that U depends on the value of q0 .
We define the electric potential V in such a manner so that it is independent
U
W
of q0 : V 

q0
q0
f
Here V  V f  Vi  V f  Vi    E  ds .
i
In all physical problems only changes in V are involved. Thus we can define
arbitrarily the value of V at a reference point, which we choose to be at infinity:
V f  V  0. We take the initial position as the generic point P with potential VP :
P
VP    E  ds . The potential VP depends only on the coordinates of P and on E.

(24-3)
Electric potential is a scalar property associated with an electric
field, regardless of whether a charged object has been placed in
that field; it is measured in joules per coulomb or volts. V
Electric potential energy is an energy of a charged object in an
external electric filed ( or more precisely, an energy of the system
consisting of the object and the external electric field); it is
measured in joules. U
W  qV
Equipotential Surfaces
A collection of points that have the same
potential is known as an equipotential
surface. Four such surfaces are shown in
the figure. The work done by E as it moves
a charge q between two points that have a
potential difference V is given by
W  qV .
For path I : WI  0 because V  0.
For path II: WII  0 because V  0.
For path III: WIII  qV  q V2  V1  .
For path IV: WIV  qV  q V2  V1  .
Note : When a charge is moved on an equipotential surface  V  0 
the work done by the electric field is zero: W  0.
(24-10)
The Electric Field E is Perpendicular
to the Equipotential Surfaces
Consider the equipotential surface at
potential V . A charge q is moved
E
F
A
q
V

r
B
S
by an electric field E from point A
to point B along a path r .
Points A and B and the path lie on S .
Let's assume that the electric field E forms an angle  with the path r .
The work done by the electric field is: W  F  r  F r cos  qE r cos .
We also know that W  0. Thus: qE r cos  0, where
q  0, E  0, r  0. Thus: cos   0    90.
The correct picture is shown in the figure below.
E
S
V
(24-11)
Examples of Equipotential Surfaces and the Corresponding Electric Field Lines
Uniform electric field
Isolated point charge
Electric dipole
Equipotential surfaces for a point charge q :
q
q
V
. Assume that V is constant  r 
 constant.
4 0 r
4 0V
Thus the equipotential surfaces are spheres with their center at the point charge
q
and radius r 
.
4 0V
(24-12)
Checkpoint 3:
The figure here shows a family or parallel equi potential surfaces ( in
cross section) and five paths along which we shall move an electron
from one surface to another.
a) What is the direction of the electric field associated with the
surfaces?
b) For each path, is the work we do positive, negative, or zero?
c) Rank the paths according to the work we do, Greatest first.
VP 
1
q
4 0 R
Definition of voltage : V  
SI Units of V :
W
q0
Units of V : J/C, known as the volt
Potential Due to a Point Charge
Consider a point charge q placed at the origin. We will use
the definition given on the previous page to determine the
potential VP at point P a distance R from O.
R



R
R
VP    E  ds   Edr cos 0   Edr
The electric field generated by q is:
q
E
4 0 r 2
O
VP 
q

dr
4 0 R r 2
dr
1


 x2 x

q  1
1 q
 VP 


4 0  r  R 4 0 R
(24-4)
Potential Due to a Group of Point Charges
q1
r1
q2
r2
q3
P
r3
Consider the group of three point charges shown in the
figure. The potential V generated by this group at any
point P is calculated using the principle of superposition.
1. We determine the potentials V1 ,V2 , and V3 generated
by each charge at point P:
V  V1  V2  V3
V1 
q1
1 q2
1 q3
, V2 
, V3 
4 0 r1
4 0 r2
4 0 r3
1
2. We add the three terms:
V  V1  V2  V3
V
q1
1 q2
1 q3


4 0 r1 4 0 r2 4 0 r3
1
The previous equation can be generalized for n charges as follows:
1 q1
1 q2
1 qn
1
V

 ... 

4 0 r1 4 0 r2
4 0 rn 4 0
n
qi
1 r
i
(24-5)
Checkpoint 4:
The figure here shows three arrangements of two protons, Rank
the arrangements according to the net electric potential produced
at point P by the protons, Greatest First
Calculating the Electric Field E from the Potential V
Now we will tackle the reverse problem, i.e., determine E if we know V .
Consider two equipotential surfaces that correspond to the values V and V  dV
separated by a distance ds as shown in the figure. Consider an arbitrary direction
represented by the vector ds . We will allow the electric field
to move a charge q0 from the equipotential surface V to the surface V  dV .
A
B
V
V+dV
The work done by the electric field is given by:
W  q0 dV (eq. 1).
Also W  Fds cos   Eq0 ds cos  (eq. 2)
If we compare these two equations we have:
dV
Eq0 ds cos   q0 dV  E cos   
.
ds
From triangle PAB we see that E cos  is the
component Es of E along the direction s.
Thus: Es  
V
.
s
Es  
V
s
(24-13)
Es  
V
s
We have proved that Es  
V
.
s
The component of E in any direction is the negative of the rate
at which the electric potential changes with distance in this direction.
A
B
V
V+dV
If we take s to be the x- , y -, and z -axes we get:
V
Ex  
x
V
Ey  
y
V
Ez  
z
If we know the function V ( x, y, z )
we can determine the components of E
and thus the vector E itself :
E
V ˆ V ˆ V ˆ
i
j
k
x
y
z
(24-14)
Potential Energy U of a System of Point Charges
We define U as the work required to assemble the
q2
y
r12
system of charges one by one, bringing each charge
from infinity to its final position.
r23
Using the above definition we will prove that for
a system of three point charges U is given by:
q1
r13
q3
x
O
q2 q3
q1q3
q1q2
U


4 0 r12 4 0 r23 4 0 r13
Note : Each pair of charges is counted only once.
For a system of n point charges
U
1
4 0
n

i , j 1
i j
qi q j
rij
.
qi  the potential energy
U is given by:
Here rij is the separation between qi and q j .
The summation condition i  j is imposed so that, as in the
case of three point charges, each pair of charges is counted only once.
(24-15)
y
Step 1

q1
W1  0
x

O
Step 2
y
q1
Step 1 : Bring in q1:
r12
q2
(no other charges around)
Step 2 : Bring in q2 :
W2  q2V (2)
V (2) 
q1
4 0 r12
 W2 
q1q2
4 0 r12
Step 3 : Bring in q3:
W3  q3V (3)
x
O
Step 3
r12
y

q2
q1
r23
r13
q3
O
x
1  q1 q2 
V (3) 
  
4 0  r13 r23 
1  q1q3 q2 q3 
W3 



4 0  r13
r23 
W  W1  W2  W3
W
qq
qq
q1q2
 2 3  1 3
4 0 r12 4 0 r23 4 0 r13
(24-16)
conductor
path
B

E 0
A
Potential of an Isolated Conductor
We shall prove that all the points on a conductor
(either on the surface or inside) have the same
potential.
A conductor is an equipotential surface.
Consider two points A and B on or inside a conductor. The potential difference
VB  VA between these two points is given by the equation:
B
VB  VA    E  d S .
A
We already know that the electrostatic field E inside a conductor is zero.
Thus the integral above vanishes, and VB  VA for any two points
on or inside the conductor.
(24-17)
Isolated Conductor in an External Electric Field
We already know that the surface of a conductor
is an equipotential surface. We also know that
the electric field lines are perpendicular to the
equipotential surfaces.
From these two statements it follows that the electric field vector E is
perpendicular to the conductor surface, as shown in the figure.
All the charges of the conductor reside on the surface and arrange
themselves in such a way so that the net electric field inside the
conductor Ein  0.
The electric field just outside the conductor is: Eout 

.
0
(24-18)
E
n̂
Eout 
E
Ein  0

nˆ
0
Electric Field and Potential
in and around a Charged
Conductor : A Summary
n̂
1. All the charges reside on the conductor surface.
2. The electric field inside the conductor is zero: Ein  0.
3. The electric field just outside the conductor is: Eout 

.
0
4. The electric field just outside the conductor is perpendicular
to the conductor surface.
5. All the points on the surface and inside the conductor have the same potential.
The conductor is an equipotential surface.
(24-19)
Electric Field and Electric Potential
for a Spherical Conductor of Radius R
and Charge q
R
1
q
.
4 0 R
For r  R ,
V
For r  R ,
1 q
V
.
4 0 r
For r  R ,
E
For r  R ,
E
1
q
.
2
4 0 R
1
q
.
2
4 0 r
Note : Outside the spherical conductor, the electric field
and the electric potential are identical to that of a point
charge equal to the net conductor charge and placed
at the center of the sphere.
(24-20)
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
Example 6:
Example 7:
Example 8:
Example 9:
Example 10:
Example 11:
Example 12:
Example 13:
Example 14:
Example 15:
Example 16:
Example 17:
Example 18:
Example 19:
Example 20:
Example 21:
Example 22:
Example 23:
Example 24:
Example 25:
Example 26:
Example 27:
Example 28: