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Chapter 29
Electric Potential
Reading: Chapter 29
1
Electrical Potential Energy
When a test charge is placed in an electric field,
it experiences a force
E
F  qE
q
s
If s is an infinitesimal displacement of test charge, then the work
done by electric force during the motion of the charge is given by
sF  qsE
2
Electrical Potential Energy
sF  qsE
E
F  qE
q
This is the work done by electric field.
s
In this case work is positive.
g
F  mg
s
This is very similar to
gravitational force: the
work done by force is
Because the positive work is done, the potential
energy of charge-field system should decrease.
So the change of potential energy is
U  sF  qsE
sF  mgs
The change of potential energy is
U  sF  mgs
sign minus
3
Electrical Potential Energy: Example
F  qE
A
q
α
s
B
UB  UA  sF  qsE  qsE cos α
E
UB  UA  qsE cos α
A
q
F  qE
α
s
UB  UA  s1F  s2F  qs1E
s1
s2
B
s2E  0
s1  s cos α
UB  UA  qsE cos α
E
The change of potential energy does not depend on the path:
The electric force is conservative
4
Electrical Potential Energy
F  qE
α
q
A
ds
s
E
ds
B
B
U  UB  UA  q  E  ds
A
ds is oriented tangent to a path
through space
For all paths:
UB  UA  qsE cos α
The electric force is conservative
5
Electric Potential
Electric potential is the potential energy per unit charge,
V
U
q
The potential is independent of the value of q.
The potential has a value at every point in an electric field
Only the difference in potential is the meaningful quantity.
UB U A
sF
VB  VA 


 sE  sE cos α
q
q
q
A
q
F  qE
α
s
B
E
6
Electric Potential
To find the potential at every point
 1. we assume that the potential is equal to 0 at some
point, for example at point A, VA  0
 2. we find the potential at any point B from the expression
B
B
A
A
VB  VA   Eds    Eds
A
VB  sE cos α
α
s
B
E
7
Electric Potential: Example
Plane: Uniform electric field
σ 0
VA  0
σ
E
ε

B
B
E
h
ds
B
σh
VB  VA   Eds  E  ds  hE  
2ε
0
A
A
A
σ 0
E
ds
h
C
C
C
σh
VC  VA   Eds  E  ds  hE  
2ε
0
A
A
8
Electric Potential: Example
V
σ 
0
h
V
σ 
Plane: Uniform electric field
B
E
h
σh
VB  
2ε
0
h
0
A
σ 
E
h
C
σh
VC  
2ε
0
All points with the same h
have the same potential
9
Electric Potential: Example
Plane: Uniform electric field
The same potential
σh
V 
2ε
0
equipotential lines
σh
V 
2ε
0
10
Electric Potential: Example
Point Charge
V  0
V  0
Q
Er  ke 2
r
ds
B
E
r
Q0
B
r
r
r
dr
1
Q
VB  VA   Eds   Er dr  keQ  2  keQ
 ke
r
r
r



11
Electric Potential: Example
Point Charge
V  0
equipotential lines
Q
Vr  ke
r
12
Electric Potential: Example
Point Charge
Q
Vr  ke
r
V  0
• The potential difference
between points A and B will be
B
1 1
VB  VA  keQ      Eds
 rB rA 
A
Q
13
Units
• Units of potential: 1 V = 1 J/C
– V is a volt
– It takes one joule (J) of work to move a 1-coulomb (C)
charge through a potential difference of 1 volt (V)
 Another unit of energy that is commonly used in atomic and nuclear
physics is the electron-volt
 One electron-volt is defined as the energy a charge-field system
gains or loses when a charge of magnitude e (an electron or a proton) is
moved through a potential difference of 1 volt
1 eV = 1.60 x 10-19 J
14
Potential and Potential Energy
• If we know potential then the potential energy of point
charge q is
U  qV
(this is similar to the relation between electric force and electric field)
F  qE
15
Potential Energy: Example
What is the potential energy of point charge q in the field of uniformly
charged plane?
U  QV
Q
σh
V 
2ε
0
h
σ
σQ  0
U
0
σQ
U  qV  
h
2ε
0
h
σQ  0
U
0
repulsion
h
attraction
16
Potential Energy: Example
What is the potential energy of two point charges q and Q?
This can be calculated by two methods:
Uq  qVQ
A
q
Q
r
Q
VQ  ke
r
UQ  QVq
B
q
Q
r
q
Vq  ke
r
The potential energy of
point charge q in the
field of point charge Q
qQ
Uq  qVQ  ke
r
The potential energy of
point charge Q in the
field of point charge q
qQ
UQ  QVq  ke
r
In both cases we have the same expression for the energy. This
expression gives us the energy of two point charges.
qQ
U  ke
r
17
Potential Energy: Example
Potential energy of two point charges:
q
Q
U  ke
r
U
qQ
r
U
qQ  0
repulsion
qQ  0
attraction
0
r
0
r
18
Potential Energy: Example
Find potential energy of three point charges:
q2
q1
r12
r13
r23
U12  ke
U  U12  U13  U23
q1q2
r12
U13  ke
q1q3
r13
U23  ke
q2q3
r23
q3
U  U12  U13  U23  ke
qq
qq
q1q2
 ke 1 3  ke 2 3
r12
r13
r23
19
Potential Energy: Applications: Energy Conservation
For a closed system: Energy Conservation:
The sum of potential energy and kinetic energy is constant
q1
q2
r12,initial
r23,initial
r13
U  U12  U13  U23
- Potential energy
mv 2
T
2
- Kinetic energy
r12,final
q3
r23,final
q2
Example: Particle 2 is released from
the rest. Find the speed of the
particle when it will reach point P.
v2
Initial Energy is the sum of kinetic energy and potential energy (velocity
is zero – kinetic energy is zero)
Einitial  T  U  T  U12  U13  U23  ke
q1q2
r12,initial
q1q3
q2q3
 ke
 ke
20
r13
r23,initial
Potential Energy: Applications: Energy Conservation
For a closed system: Energy Conservation:
The sum of potential energy and kinetic energy is constant
q1
q2
r12,initial
r23,initial
r13
r12,final
q3
r23,final
q2
v2
Final Energy is the sum of kinetic energy and potential energy (velocity
of particle 2 is nonzero – kinetic energy)
Efinal  T  U  T  U12  U13  U23
qq
qq
m2v 22
qq

 ke 1 2  ke 1 3  ke 2 3
2
r12,final
r13
r23,21final
Potential Energy: Applications: Energy Conservation
For a closed system: Energy Conservation:
q1
The sum of potential energy and kinetic
energy is constant
q2
r12,initial
r23,initial
r13
r12,final
q2
Final Energy = Initial Energy
q3
ke
q1q2
r12,initial
r23,final
v2
q1q3
q2q3
q1q3
q 2 q3
m2v 22
q1q2
 ke
 ke

 ke
 ke
 ke
r13
r23,initial
2
r12,final
r13
r23,final
q2q3
q2q3
m2v 22
q1q2
q1q2
 ke
 ke
 ke
 ke
2
r12,initial
r12,final
r23,initial
r23,final
v2 
2
m2

 1
1
k
q
q

 e 1 2 

 r12,initial r12,final

 1
1

k
q
q

 e 2 3 

 r23,initial r23,final

 
 
22
Electric Potential: Continuous Charge Distribution
23
Electric Potential of Multiple Point Charge
Q
Vr  ke
r
P
q4
q1
q2
q3
q3
q1
q2
q4
Vr  V1  V2  V3  V4  ke  ke
 ke
 ke
r1
r2
r3
r4
The potential is a scalar sum.
The electric field is a vector sum.
24
Electric Potential of Continuous Charge Distribution
• Consider a small charge element dq
– Treat it as a point charge
• The potential at some point due to this
charge element is
dq
dV  ke
r
• To find the total potential, you need to
integrate to include the contributions
from all the elements
V  0
dq
V  ke 
r
The potential is a scalar sum.
25
The electric field is a vector sum.
Spherically Symmetric Charge Distribution
Uniformly distributed charge Q
Q
V ?
26
Spherically Symmetric Charge Distribution
Two approaches:
“Complicated” Approach A:
dq
V  ke 
r
“Simple” Approach B:
B
VB  VA   Eds
(simple - only because we know E(r))

27
Spherically Symmetric Charge Distribution
Q
Er  ke 3 r
a
r a
Q
Er  ke 2
r
r a
28
Spherically Symmetric Charge Distribution
V  0
r a
V  0
ds
Q0
B
E
Q
Er  ke 2
r
r
B
r
r
r
dr
1
Q
VB  VA   Eds   Er dr  keQ  2  keQ
 ke
r
r
r



29
Spherically Symmetric Charge Distribution
V  0
V  0
r a
Er  ke
ds
Q
r
a3
r
Q
Er  ke 2
r
E
C
B
Q0
B

a


r
r
a
VB  VA   Eds   Er dr   Er dr   Er dr 
a

2


Q
dr
Q
Q
Q
r
2
2
 ke 3  rdr keQ  2  ke 3 (a  r )  ke  ke
3  2 
a r
r
2a
a
2a 
a 30
r
Spherically Symmetric Charge Distribution
Vr  ke
Q
r
Q
r2 
VB  ke
3  2 
2a 
a 
r a
r a
31
Important Example
    
     
    
     
E 0
σ
E
ε

E 0
σ 0
E
σ 0
σ 0
V ?
σ 0
32
Important Example
A
VA  0
ds
B
B
E 0
σ
E
ε

E 0
VB    Eds  0
A
σ 0
E
σ 0
33
Important Example
A
VA  0
ds
E 0
σ 0
C
σ
E
ε

x
h
E
B
σ 0
E 0
B
C
B
B
A
A
C
C
VB    Eds    Eds   Eds    Eds  Ex
34
Important Example
A
VA  0
ds
E 0
σ 0
C
σ
E
ε

E
h
D
σ 0
E 0
B
B
C
D
B
D
A
A
C
D
C
VB   Eds   Eds   Eds   Eds   Eds  Eh
35
Important Example
V 0
E 0
σ
E
E
ε

σ 0
x
σ 0
E 0
σ
V  x
ε

σ
V  Eh   h
ε

V
h
x
σ
 h
ε
36
q2
q1
r12
r13
U  U12  U13  U23
r23
q1q3
q2q3
q1q2
 ke
 ke
 ke
r12
r13
r23
q3
r a
Q
Vr  ke
r
r a
Q
r2 
VB  ke
3  2 
2a 
a 
37