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Chapter 21
Electric Potential
Conservative force
Conservation of energy
charged
Potential (energy)
+ + + +
Potential
difference
(voltage)
air → conductor
discharge
2
Coulomb force is conservative
Comparing :
m1m2
F G 2
~
r
Q1Q2
F k 2
r
Electrostatic / Coulomb force is conservative
Work doesn’t depend on the path:
W   qE  dl 
L1
 qE  dl

(  E  0 )
 E  dl
0
L2
The circulation theorem of electric field
3
Electric potential
Conservative force → potential energy U
U depends on the charge!
Define electric potential:
V U / q
+
-
Potential difference / voltage:
U a  U b Wab
Vab  Va  Vb 

q
q
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Electric potential & electric field
Difference in potential energy between a and b:
b
Ub  U a   qE  dl
a
b
 Vba  Vb  Va   E  dl
a
b
a
If position b is the zero point:
Va  
V 0
a
E  dl
Relationship between electric field and potential
Unit of electric potential: Volt (V)
5
Zero point
Electric potential:
Va  
V 0
a
E  dl
It depends on the chosen of zero point
Usually let V = 0 at infinity:

Va   E  dl
a
For the field created by a point charge :
Va  

a
Q
4 0 r
2
 dr 
.a
Q
4 0 ra
Q
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Properties of electric potential
Va  
V 0
a
E  dl
b
Vba  Vb  Va   E  dl
a
1) Va → potential energy per unit (+) charge
2) Va depends on the zero point, Vba does not
3) Va and Vba are scalars, differ from E vector
U a  qVa  q 
V 0
a
E  dl ,
b
Wab  qVab  q  E  dl
a
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Calculation of electric potential
If the electric field is known:
Va  
V 0
a
E  dl
Potential of point charge:
V
Q
4 0 r
(V = 0 at infinity)
System of several point charges:
V 
Qi
4 0 ri
or
V 
dQ
A. r
dQ
4 0 r
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Potential of electric dipole
Example1: (a) Determine the potential at o, a, b.
(b) To move charge q from a to b, how much
work must be done?
Solution: (a) Vo 
Vb 
(b)
Q
4 0l

Q
8 0l

Vba  Vb  Va 
Q
4 0 r

Q
4 0 r
.a
Q
8 0l
Q
8 0l
 0 , Va  0
Q
l.
o
h
l
Q
.b
Qq
 W  qVba 
8 0l
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Charged conductor sphere
Example2: Determine the potential at a distance r
from the center of a uniformly charged conductor
sphere (Q, R) for (a) r>R; (b) r=R; (c) r < R.
Solution: All charges on surface:
E
Q
4 0 r 2
(r  R), E  0 (r  R)
Va  
r
4 0 r
2
dr 
Q
4 0 r
a
.
+
Q
+
Q
+
+
+
(a) V at r>R:

+
+
+
+
+
Same as point charge
10
(b) V at r=R:
+

Q
R
4 0 r 2
Vb  
dr 
Q
+
4 0 R
Vc   E  dl  Vb  Vb 
c
Q
+
Q
4 0 R
+
.c
+
(c) V at r<R:
b
+ .b
+
+
+
+
Same potential at
any point!
Discussion:
(1) Conductor: equipotential body
(2) Charges on holey conductor
++ +
-
-
-
?
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Uniformly charged rod
Example3: A thin rod is uniformly charged (λ, L).
Determine the potential for points along the line
outside of the rod.
Solution: Choose infinitesimal charge dQ
Va 

d L
d
 dx
4 0 x

dL

ln
4 0
d
Potential at point b?

.b dx
x
a
dQ
L
d
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Charged ring
Example4: A thin ring is uniformly charged (Q, R).
Determine the potential on the axis.

Solution: Va 
Ring
or:
Va 
dQ
4 0 r

Q
.a
4 0 r
Q
Q
4 0 x 2  R 2
r
x
dQ
R
Discussion:
(1) Not uniform?
(2) x >> R
(3) other shapes
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Conical surface
Question: Charges are uniformly distributed on a
conical surface (σ, h, R). Determine the electric
potential at top point a. (V = 0 at infinity)
a
h
R
14
Electrostatic induction
Question: Put a charge q nearby a conductor sphere
initially carrying no charge. Determine the electric
potential on the sphere.
- Q
R
+
+ o
+
+
+
+
q
b
Connect to the ground?
15
*Breakdown voltage
Air can become ionized due to high electric field
Air → conducting → charge flows
Breakdown of air: E  3 106 V / m
Breakdown voltage for a spherical conductor?
V
Q
4 0 R
 ER  R
R  5cm  V  15000V
R
Q
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Equipotential surfaces
Visualize electric potential: equipotential surfaces
Points on surface → same potential
(1) Surfaces ⊥ field at any point.
(2) Move on surface, no work done
(3) Move along any field line, V ↘
(4) Surfaces dense → strong field
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Some examples
(1) Equipotential surfaces for dipole system:
(2) Conductors
(3) Parallel but not uniform field lines?
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E determined from V
To determine electric field from the potential, use
b
Vb  Va   E  dl
a
In differential form:
dV  E  dl   El dl
Component of E in the direction of dl :
El  dV / dl
Total electric field:
V
V
V
E
i
j
k
x
y
z
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Gradient of V
dV
dV
El  
E
dl
dn
V+dV
n
V
dV
Gradient of V : V 
n
dn
dn
a
dl
b
E
Gradient → partial derivative in
direction V changes most rapidly
V
V
V
E  V  
i
j
k
x
y
z

1
2

V   E  dl
a
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Determine E from V
Example5: The electric potential in a region of
space varies as V = x2yz. Determine E.
Solution:
V
 2 xyz ,
Ex  
x
V
  x 2 z,
Ey  
y
V
  x2 y
Ez  
z
 E  2 xyzi  x zj  x yk
2
2
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Determine V from E
Example6: The electric field in space varies as
E  2 xyi  ( x 2  y 2 ) j . Determine V.
Solution:
V
Ex  
 2 xy
x
 V   2 xydx   x y  C ( y )
2
1 3
y C
3
V
dC ( y )
2
 x2  y 2
Ey  
x 
y
dy
1 3
2
( C is constant )
 V  x y  y  C
3
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