PHYS 2421
Fields and Waves
Instructor: Jorge A. López
Office: PSCI 209 A, Phone: 747-7528
Textbook: University Physics 11e, Young and Freedman
23.1 Electric potential energy
23.2 Electric potential
23.3 Calculating electric potential
23.4 Equipotential surfaces
23.5 Potential gradient
Potential varies in space
This variation can be represented with “equipotential” surfaces
An analogy:
Equal altitude lines
Some examples
Notice that E-field lines
are perpendicular to
equipotential surfaces
•All points on an equipotential surfaces are at same V
•Magnitude of E-field is NOT constant on equipotential surfaces
Free charges move perpendicular to equipotential
surfaces never along a constant V surface
Hwk: Problem 23.44 (11th Ed.) or 23.49 (12th Ed.)
Another example
A charged plate
A charged sphere
Combine into
Hwk: Problem 23.47 (11th Ed.) or 23.45 (12th Ed.)
All points in a conductors are at the same potential
Reasoning:
• Charged conductors have E perpendicular
to surface outside: E0 and E=0
• Since equipotential surfaces must be  to E
 equipotential surface must be  to surface of conductor
•  Surface of conductor is at a constant potential
• Since E=0 inside  whole conductor is at the same potential
All points in a conductors are at the same potential
Summary of Section 23.4
•E-field lines are  to
equipotential surfaces
•Points on equipotential
surfaces have same V
• Equipotential surfaces
must be  to E
•Magnitude of E-field is NOT
constant on equipotential surfaces
Hwk Section 23.4: Problems 23.44 and 23.47 (11 th Ed.)
or 23.45 and 23.49 (12th Ed.)
a
Inverting Va  Vb   E  dl to get E from V:
b
b

b
b


a
a
Take Va  Vb   dV   dV  E  dl
a
 dV  E  dl
V
V
V
 dV  Ex dx  Ey dy  Ez dz but dV 
dx 
dy 
dz
x
y
z
V
V
V
 Ex  
, Ey  
, Ez  
x
y
z
One can obtain the components of E
from the partial derivatives of V!
 V ˆ V ˆ V
i
j
y
z
 x
That is E   

ˆ
k  or E  V

Where the gradient is
In spherical coordinates E  
r
 f
f ˆ f ˆ 
f   iˆ 
j k

x

y
z 

V
r
Let us look at some examples
Since V is V 
In cartesian V 
q
the field is
4 0 r
  q / 4 0 r 
V
Er  

r
r
q  1/ r
q  1/ r 


4 0 r
4 0

q
4 0 x  y  z
2

2
2
q
4 0 r
2
E
q
4 0 r
2
rˆ

2
2
2

1/
x

y

z
V
q
q
x
Ex  


x
4 0
x
4 0  x 2  y 2  z 2 3/ 2
q
 x ˆ y ˆ z ˆ
E
i  j  k 
rˆ
2 
2
4 0 r  r
r
r  4 0 r
q
Same result !
Hwk: Problem 23.41 (11th Ed.) or 23.47 (12th Ed.)
2

Since
V
the field is
Q
4 0 x 2  a 2
V
Ex  
x


2
2

1/
x

a
V
Q
Q
x
Ex  


x
4 0
x
4 0 x 2  a 2


3/ 2
Same result as in section 21.5 !
Hwk: Problem 23.84 parts a, b and c only (11 th Ed.)
or 23.86 parts a, b and c only (12th Ed.)
Summary of Section 23.5
 V ˆ V ˆ V
E  
i
j
E in terms of V:
y
z
 x

ˆ
k

V
V
V
Ex  
, Ey 
, Ez 
x
y
z
In spherical coordinates
V
Er  
r
Homework for Section 23.5:
Problems (11th Ed.) 23.41 and 23.84 a, b and c
or (12th Ed.) 23.47 and 23.86 a, b and c
Chapter 23:
Electric Potential
23.1 Electric potential energy
23.2 Electric potential
23.3 Calculating electric potential
23.4 Equipotential surfaces
23.5 Potential gradient