
Flashback to PHY113…
 Work done by a conservative force is independent of the path of the object.
 Gravity and elastic forces are examples.
 This leads to the concept of potential energy and helps us avoid tackling problems using only forces.
xf
W   Fs dx  U
xi

Electrostatic forces are also conservative.
B

E


F  q0 E
 
U    F.d s  q0
path
 
 E.d s
path
A
q0
…but F is a conservative force
…so the path we take
does not matter
 
U  q0  E.d s
B
A
Work, U, is dependent on the magnitude of the test charge, q0.
 We’d like to have a quantity independent of the test charge and only an attribute of the electric field.

U
V
q0
 
U
V 
   E.d s
q0
A
Electric Potential
(J/C=V)
Potential Difference
between A and B
B
P
VP    E.ds

Electric Potential at P
Potential
V J C
Energy
J  N m
Electric Field
N  V .C  J 
N C 
 V m

C  J  N .m 
Electron-Volt
1eV  e1V   1.6 10 19 C 1 J C   1.6 10 19 J


B
B
B
A
A
A
V    E.ds    E cos 0ds   E  ds
V   Ed
U  q0 Ed
Electric field lines point to decreasing
potential.
A positive charge will lose potential energy
and gain kinetic energy when moving in the
direction of the field.
D
B
s
E
B
B
A
A
V    E.ds  E. ds  E.s

C
A
VAB  (E.s) AC  (E.s) CB
U  q0 (E.s) AC  q0 (E.s) CB
VAB   Es AC cos 0  EsCB cos 90
U  q0 Es AC cos 0
VAB   Es cos 
U  q0 Es cos 
D
D
A
A
VAD    E.ds   E. ds   E.s  ( E.s) AC  ( E.s)CD   Es cos 
No work is done moving a charged particle perpendicular to
a field (along equipotential surfaces)
Uniform Field
Point Charge
Electric Dipole
B
VB  VA    E.ds
A
q
q
q
ˆ
E.ds  ke 2 r.ds  ke 2 ds cos   ke 2 dr
r
r
r
VB  VA    Er dr
1 1
dr
k
q

 
e 
2
r
r
 B rA 
rA
rB
VB  VA  ke q 
V A   0
q
V  ke
r
 q1 q2 
VP  ke   
 r1 r2 
q1q2
U  ke
r12
 q1q2 q2 q3 q3q1 



U  ke 
r23
r31 
 r12
Two test charges are brought separately into the vicinity of
a charge +Q. First, test charge +q is brought to a point a
distance r from +Q. Then this charge is removed and test
charge –q is brought to the same point. The electrostatic
potential energy of which test charge is greater:
1. +q
2. –q
3. It is the same for both.
E
B
V    E.ds
A
E
dV
Er  
dr
dV  E.ds
V
In general: E x  
x
dV
Ex  
dx
V
Ey  
y
V
Ez  
z
2ke qa
VP  2
2
x a
2ke qa
V
x2
(x >> a)
dV 4ke qa
Ex  

dx
x3
Between the Charges
2ke qx
VP  2
a  x2
 a2  x2 
dV

Ex  
 2 k e q  2
2 
2

dx
 a  x  
Start with an infinitesimal
charge, dq.
dq
dV  ke
r
Then integrate over the
whole distribution
dq
V  ke 
r
keQ
V
E
x2  a2
x
keQx
2
a

2 3/ 2
keQ  l  l 2  a 2
ln
V

l
a





Er  k e
Q
r2
r
r
dr
r2

VB    Er dr  keQ 
r>R
kQ
Er  e 3 r
R
r<R
r
VD  VC    Er dr 
R
keQ 
r2 
 3  2 
VD 
2R 
R 

keQ 2 2
R r
3
2R


VB  ke
Q
r
VC  k e
Q
R






Charges always reside at the outer surface of the conductor.
The field lines are always perpendicular to surface.
Then E.ds=0 on the surface at any point.
Which means, VB‐VA=0
along the surface.
The surface is an equipotential surface.
Finally, since E=0 inside the conductor, the potential V is constant and equal to the surface value.
q1 r1

q2 r2
E1 r2

E2 r1
B
VB  VA    E.ds  0
A
We can always find a
path where E.ds is
non-zero.
But, since V=0 for all
paths, E must be zero
everywhere in the cavity.
A cavity without any charges enclosed by a
conducting wall is field free.
Charges can have different electric potential energy at different points in an electric field.
 Electric potential is the electric potential energy per unit charge.
 All points inside a conductor are at the same potential.


Reading Assignment
 Chapter 26 – Capacitance and Dielectrics

WebAssign: Assignment 3