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Physics 202, Lecture 5
Today’s Topics
  Electric Potential Energy, Electric Potential (Ch. 23)
  Motion of charges in E fields
  Methods for obtaining E, V
  Point charges (review)
  Continuous distributions
Electric Potential and Electric Field Lines
+q
-q
higher V
E
lower V
Field lines always point towards lower electric potential!
In an electric field:
  positive charges are always subject to a force in the
direction of field lines, towards lower V
  negative charge is always subject a force in the
opposite direction of field lines, towards higher V
Electric Potential - Uniform Field
B
 
 
+ ΔV = − E • dl ⇒ VB − VA = ∫ − E • dl
B
B
A
 
= − ∫ E dl = −E ∫ dl = −Ex
A
 
E || dl
Recall:
B
A
x
A
ΔV
VB − VA
E=−
=−
x
x
E constant
ΔU = qΔV
A positively charged particle moving from A
to B gains kinetic energy K = potential
energy ΔU lost by the charge-field system
ΔK + ΔU = 0
(energy conservation)
Example: Uniform Electric Field
In the uniform electric field shown:
1. What is the potential at B,C,D,G?
2. If a charge +q is placed at B,
what is the potential energy UB?
D
3. If now a –q is at B, what is UB?
A
VA=0
C
B
G
d
4. If a -q is initially at rest at G,
will it move to A or B? What is its final kinetic energy?
Particles will move to minimize their final potential energy
See also text: 23.3,…
Cathod Ray tube
Summary: Methods for Obtaining E, V

q 
E = ∑k 2 r
r
Coulomb’s Law:
(lecture 2)

dq
E = k ∫ 2 rˆ
r
€
 
V = − ∫ E i dl
  qencl
∫ E idA = ε 0
Gauss’s Law:
(lecture 3,4)
Potential:
(lecture 4,today)
€
q
V = ∑k
r
dq
V = k∫
r

E = −∇V
∂V
∂V
Ex = − ∂V
,
E
=
−
,
E
=
−
∂x y
∂y z
∂z
Know all methods and when to apply which
Electric Potential and Point Charges
For point charge q shown below, what is VB-VA ?
B
B
 1 1
dr
VB − VA = − ∫ E(r)dr = −kq ∫ 2 = kq  − 
r
 rB rA 
A
A
B
rB
A
rA
q
independent of path b/w A and B!
Potential of point charge:
kq
V (r) =
r
Many point charges: superposition
qi
V (r) = k ∑
i ri
Continuous Charge Distributions: “Brute Force”
Finite charge distributions: usually set V=0 at infinity
(can also set it zero at ground)
If charge distribution is known:
dq
dV = k
r
r=
V = ∫ dV
distance b/w source and obs. point
Note: scalar integral
Examples:
finite line (23.39), ring (23.33), disk…
€
Potential due to Ring
 Imagine a homogenously charged ring
V=
∫ dV =k ∫
V =k∫
dq
r
dq
2
x +R
2
=k
1
2
x +R
2
∫ dq = k
Q
x 2 + R2
Continuous Charge Distributions:Gauss’s Law
Obtain E by Gauss’s Law.
Example 1:
spherical shell
Integrate to get V.
ΔV = − ∫
B
A
 
E i dl = VB − VA
V=constant inside E=0 region
Er =
E =0
Note: like spherical conductor
r<R
kQ
r2
r>R
Continuous Charge Distributions:Gauss’s Law
Uniformly charged sphere:
Text example: 23.19
R
Use
ΔV = − ∫
B
A
 
E i dl = VB − VA
since from Gauss’s law:
E=
kQr
R3
r<R
E=
kQ
r2
r>R
Other examples:
23.23, parallel plate,…
Compute E from V
 Imagine an electric potential of the following form
2
2
V (x, y,z) = 2x + 8y z + 2z
∂V
Ex = −
∂x
€
∂V
, Ey = −
∂y
∂V
,Ez = −
∂z
∂V
Ex = −
= −4 x
∂x
∂V
Ey = −
= −16zy
∂y
∂V
Ez = −
= −8y 2 − 4z
∂z
or
2
Units of V

E = −∇V
Units of V/m
1/--pages
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