Electric Potential Energy
Work done by Coulomb force when q
1 moves from a to b: r q
1
(+) dr ds a r a q
1
F
(+) r b b q
2
(-)

Electric Potential Energy
The important point is that the work depends only on the initial and final positions of q
1
.
r a a r dr ds
F
E r ab b q
1
(+) q
2
(-)
In other words, the work done by the electric force is independent of path taken . The electric force is a conservative force.

Electric Potential Energy
A charged particle in an electric field has electric potential energy.
It “feels” a force (as given by
Coulomb’s law).
It gains kinetic energy and loses potential energy if released. The
Coulomb force does positive work, and mechanical energy is conserved.
+ + + + + + + + + + + + + +
+
F
E
- - - - - - - - - - - - - - - - - - -

Electric Potential
Dividing W by Q gives the potential energy per unit charge.
V
AB
B.
, is known as the potential difference between points A and
The electric potential V is independent of the test charge q
0
.

Electric Potential
If V
AB is negative, there is a loss in potential energy in moving is being done by the field.
if it
Q from A to B; the work is positive, there is a gain in potential energy; an external
agent performs the work
+ + + + + + + + + + + + + +
+
F
E
- - - - - - - - - - - - - - - - - - -

Electric Potential
V
AB is the potential at B with reference to A
V
B and and A
V
A are the potentials (or absolute potentials) at B

Electric Potential
If we choose infinity as reference the potential at infinity is zero;the electric potential of a point charge q is
  
1 q
4

0 r
.
The potential at any point is the potential difference between that point and a chosen point in which the potential is zero.

Things to remember about electric potential:
 Electric potential difference is the work per unit of charge that must be done to move a charge from one point to another without changing its kinetic energy.
● Sometimes it is convenient to define V to be zero at the earth
(ground).
 The terms “electric potential” and “potential” are used interchangeably.
 The units of potential are joules/coulomb:
1 joule
1 volt =
1 coulomb

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.
y
3 m
P q
1
4 m q
2 x
V = k
P
 i r i q i = k

 q q
1 + r r
1 2
2


= 9×10
9


1×10
-6
-4×10
+
3 5
-6


Thanks to Dr. Waddill for the use of these examples.

Example: how much work is required to bring a +3  C point charge from infinity to point P?
y q
3
P
3 m
W external

3

P

V


0
W external
  
6

  3
 x q
1
4 m q
2
W external
  
3
1.26 10 J
The work done by the external force was negative, so the work done by the electric field was positive. The electric field “pulled” q
3 in (keep in mind q
2 is 4 times as big as q
1
).
Positive work would have to be done by an external force to remove q
3 from P.

Electric Potential of a Charge Distribution
Collection of charges:
V
P

1
4

0
 i q i .
r i
P is the point at which V is to be calculated, and r i charge from P.
is the distance of the i th
Charge distribution:
V

1
4

0
 dq r
.
Potential at point P.
P r dq

Electric Potential of a Charge Distribution

Example: A rod of length L located along the x-axis has a total charge Q uniformly distributed along the rod. Find the electric potential at a point P along the y-axis a distance d from the origin.
y
P d r x
L dq dx x
 =Q/L dq=  dx dV
 k dq r
 k
  L
V dV
0
 dx x
2  d
2
Thanks to Dr. Waddill for this fine example.

y
P d r x
L dq dx
V
 
0
L k
 dx x
2  d
2
 k
Q
L

0
L dx x
2  d
2 x

A good set of math tables will have the integral: dx x
2  d
2


 x
2  d
2

V
 kQ L L d
L ln


 d
2  2



Example: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring.
dQ
R r
P x
Every dQ of charge on the ring is the same distance from the point P.
x dV
 k dq r
 k dq x
2 
R
2
V
  ring dV
 k
 ring dq x
2 
R
2

dQ
R x r
P x
V
 k x
2 
R

2 ring dq
V
 kQ x
2 
R
2

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.
R r dQ x
P x
The disc is made of concentric rings. The area of a ring at a radius r is 2  rdr, and the charge on each ring is  (2  rdr).
We *can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.
dV ring
 x
2  r
2

dQ r
P x x
R
V
  ring dV

1
4

0
 ring x
2  r
2


2

0

0
R rdr x
2  r
2
V


2

0
R x
2  r
2 

2

0

0 x
2 
R
2  x


Q
2

0
R
2
 x
2 
R
2  x

 
Q

R
2

dQ
R r x
P x
V

Q
2

0
R
2
 x
2 
R
2  x

Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 3?

MAXWELL'S EQUATION
• The line integral of E along a closed path is zero
• This implies that no net work is done in moving a charge along a closed path in an electrostatic field

MAXWELL'S EQUATION
• Applying Stokes's theorem
• Thus an electrostatic field is a conservative field

Electric Potential vs. Electric Field
• Since we have
V
• The negative sign shows that the direction of E is opposite to the direction in which V increases

Electric Potential vs. Electric Field
• If the potential field V is known, the E can be found

Example: In a region of space, the electric potential is V(x,y,z)
= Axy 2 + Bx 2 + Cx, where A = 50 V/m 3 , B = 100 V/m 2 , and C =
-400 V/m are constants. Find the electric field at the origin
E (0, 0, 0) x
 

V
 x
(0,0,0)
 

Ay
2 
2Bx

C

(0,0,0)
 
C
E (0, 0, 0) y
 

V
 y
(0,0,0)
 
(2Axy)
(0,0,0)

0
E (0, 0, 0) z
 

V
 z
(0,0,0)

0
E(0,0,0)

400
V ˆ m i