ELECTRIC POTENTIAL
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 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)
Picture a Region of
space Where there is an Electric
Field
• Imagine there is a particle of charge q at
some location.
• Imagine that the particle must be moved to
another spot within the field.
• Work must be done in order to
accomplish this.
So, when we move a charge in an
Electric Field ..


Move the charge at constant velocity so it is
in mechanical equilibrium all the time.
Ignore the acceleration at the beginning
because you have to do the same amount of
negative work to stop it when you get there.

When an object is moved from one point to
another in an Electric Field,



It takes energy (work) to move it.
This work can be done by an external force (you).
You can also think of this as the
FIELD
negative of
doing the
this amount of work on the particle.
The net work done by a conservative (field)
force on a particle moving
around a closed path is
ZERO!
IMPORTANT

The work necessary to move a charge
from an initial point to a final point is
INDEPENDENT OF THE PATH
CHOSEN!
Electric Potential Energy



When an electrostatic force acts between
two or more charged particles, we can
assign an ELECTRIC POTENTIAL ENERGY
U to the system.
The change in potential energy of a charge
is the amount of work that is done by an
external force in moving the charge from its
initial position to its new position.
It is the negative of the work done by the
FIELD in moving the particle from the initial
to the final position.
Definition – Potential Energy


PE or U is the work done by an external
agent in moving a charge from a
REFERENCE POSITION to a different
position.
A Reference ZERO is placed at the most
convenient position

Like the ground level in many gravitational
potential energy problems.
Example:
Work by External Agent
Wexternal = F  d = qEd= U
E
d
Zero Level
q
F
Work done by the Field
is:
Wfield= -qEd
= -Wexternal
AN IMPORTANT DEFINITION

Just as the ELECTRIC FIELD was defined as
the FORCE per UNIT CHARGE:
F
E
q
VECTOR
We define ELECTRICAL POTENTIAL as the
POTENTIAL ENERGY PER UNIT CHARGE:
U
V
q
SCALAR
UNITS OF POTENTIAL
U
Joules
V 
 VOLT
q Coulomb
Furthermore…
U Wapplied
V 

q
q
so
Wapplied  qV
If we move a particle through a potential
difference of V, the work from an external
“person” necessary to do this is qV
Consider Two Plates
OOPS …
Look at the path issue
An Equipotential Surface is defined as a
surface on which the potential is constant.
V  0
It takes NO work to move a charged particle
between two points at the same potential.
The locus of all possible points that require NO
WORK to move the charge to is actually a surface.
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
Lets 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
q  0, E  0, r  0 Thus: cos   0    90
The correct picture is shown in the figure below
E
S
V
(24 - 11)
Field Lines are Perpendicular to the
Equipotential Lines
Equipotential
Work external  q0 (V f  Vi )
Keep in Mind
W  F d


Force and Displacement are
VECTORS!
Potential is a SCALAR.
UNITS




1 VOLT = 1 Joule/Coulomb
For the electric field, the units of N/C can be
converted to:
1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM)
Or
1 N/C = 1 V/m

So an acceptable unit for the electric field is now
Volts/meter. N/C is still correct as well.
In Atomic Physics



It is sometimes useful to define an energy
in eV or electron volts.
One eV is the additional energy that an
proton charge would get if it were
accelerated through a potential
difference of one volt.
1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6
x 10-19 Joules.
Coulomb Stuff:
A NEW REFERENCE: INFINITY
Consider a unit charge (+) being brought from
infinity to a distance r from a
Charge q:
q
x
r
1
q
E
2
40 r
To move a unit test charge from infinity to the point
at a distance r from the charge q, the external force
must do an amount of work that we now can
calculate.
VB  VA    E  ds
q
   k 2 runit  ds
r
rB
1 1
dr
 kq  2  kq  
r
 rB rA 
rA
Set the REFERENCE LEVEL OF POTENTIAL
at INFINITY so (1/rA)=0.
For point charges
qi
V

40 i ri
1
Potential due to a group of point charges
Consider the group of three point charges shown in the
q1
r1
q2
r2
P
r3
q3
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 V 3 generated
by each charge at point P.
V  V1  V2  V3
V1 
q1
1 q2
1 q3
, V2 
, V3 
4 o r1
4 o r2
4 o r3
1
2. We add the three terms:
V  V1  V2  V3
V
q1
1 q2
1 q3


4 o r1 4 o r2 4 o r3
1
The previous equation can be generalized for n charges as follows:
q1
1 q2
1 qn
1
V

 ... 

4 o r1 4 o r2
4 o rn 4 o
1
n
qi
1 r
i
(24 - 5)
For a DISTRIBUTION of charge:
dq
V   k
r
volume
Example : Potential created by a line of charge of
length L and uniform linear charge density λ at point P.
Consider the charge element dq   dx at point A, a
distance x from O.
From triangle OAP we have:
r  d 2  x 2 Here d is the distance OP
The potential dV created by dq at P is:
1 dq
1
 dx
dV 

4 o r
4 o d 2  x 2
O
dq
A

V
4 o

L

0
dx
d 2  x2

V
4 o

V
4 o
dx
d 2  x2

 ln x  d 2  x 2

ln L 
 


L  x   ln d 

L
ln x  d 2  x 2 

 0
2
2
(24 - 8)
Calculating the electric field E from the potential V
(24 - 13)
The work done by the electric field is given by:
W  qo dV (eqs.1)
V
A
also W  Fds cos   Eqo ds cos  (eqs.2)
B
If we compare these two equations we have:
dV
Eqo ds cos   qo dV  E cos   
ds
From triangle PAB we see that E cos  is the
V+dV
component Es of E along the direction s.
Thus: Es  
V
s
Es  
V
s
Es  
V
s
We have proved that: Es  
V
s
(24 - 14)
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 ro 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 - 15)
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
Using the above definition we will prove that for
r23
q1
a system of three point charges U is given by:
r13
q3
x
O
q2 q3
q1q3
q1q2
U


4 o r12 4 o r23 4 o r13
Note : each pair of charges is counted only once
For a system of n point charges
U
1
4 o
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
y
Step 1
(24 - 16)

q1
x

O
Step 2
y
q1
r12
q2
Step 1 : Bring in q1
W1  0
(no other charges around)
Step2 : Bring in q2
W2  q2V (2)
V (2) 
q1
4 o r12
 W2 
q1q2
4 o r12
Step3 : Bring in q3
W3  q3V (3)
x
O
Step 3

q2
r12
y
q1
r23
r13
q3
O
1  q1 q2 
  
4 o  r13 r23 
1  q1q3 q2 q3 
W3 



4 o  r13
r23 
V (3) 
x
W  W1  W2  W3
q2 q3
q1q3
q1q2
W


4 o r12 4 o r23 4 o r13
(24 - 17)
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
conductor
path
B

E 0
A
A conductor is an equipotential surface
Consider two points A and B on or inside an conductor. The potential difference
VB  VA between these two points is give 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.
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 as way so that the net electric field inside the
conductor Ein  0.
The electric field just out side the conductor is: Eout 


(24 - 18)
(24 - 19)
E
n̂
Eout

 nˆ
o
Electric field and potential
in and around a charged
conductor. A summary
E
Ein  0
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 

o
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 eequipotential surface
In the figure, point P is at the center of the
rectangle. With V = 0 at infinity, what is the net
electric potential in terms of q/d at P due to the six
charged particles?
Continuing
1
2
3
5
6
s
4
2
d 2 5d 2
d 
2
2
2
s  d    d 

2
4
4
 
d
s
5  1.12d
2
qi
  2q  2q 3q  3q  5q  5q 
 k


r
d
/
2
1
.
12
d


i
i
q
q
  8q 10q 


V  k


k

8

8
.
93

0
.
93
k

d
1
.
12
d
d
d


q
V  8.35 x109
d
V  k