Chapter 23--Examples
1
Problem

In the figure below, point P is at the
center of the rectangle. With V=0 at
infinity, what is the net electric potential
at P due to the six charged particles?
+5q
d
-2q
d
+3q
d
d
P
d
-2q
-3q
d
+5q
2
Find distance from corners to P
d
d
-2q
+5q
s
d/2
d
+3q
d
P
d
-3q
d
d
-2q
+5q
3
Potentials (Voltage) is a scalar
5q
V3  k
s
2
3
d
+5q
V2  k
-2q
 2q
d
2
d
+3q
4
-2q
5
V1  k
 3q
s
s
P
 2q
V5  k
d
2
-3q
d
d/2
d
3q
V4  k
s
1
d
d
d
+5q
5q
V6  k
s
6
V  Vi  V1  V2  V3  V4  V5  V6
i
4
V  Vi  V1  V2  V3  V4  V5  V6
i
V  V1  V2  V3  V4  V5  V6
  3q  2q 5q 3q  2q 5q 
V  k


 
 
 s
d
d
s
s
s 
2
2


  4q 10q 

V  k

 d
s 
 2

kq
V
4 5 8
d
0.94kq
V
d


5
Problem
A charge q is distributed uniformly throughout a
spherical volume of radius, R.
a)
Setting V=0 at infinity shown that the potential at
a distance r from the center, where r<R is given
by

q 3R  r
V
80 R 3
b)
2
2

What is the potential difference between a point
on the surface and the sphere’s center?
6
First, use Gauss’s law to find the Efield inside and outside the sphere
 qenclosed

 Eoutside  dA 
0


Eoutside 4r 

Eoutside 
2
q
q
0
1
rˆ
2
40 r
 qenclosed

 Einside  dA 
0


Einside 4r 2 

1
0
 V 
q
4 3
R
3


1  4 3
  r 
0  3



q
Einside 
rrˆ 
rrˆ
3
3 0
40 R
Einside 4r 2 
7
Outside is simpler

q 1
Eoutside 
rˆ
2
40 r
R 
R
 
q 1
V f  Vi    E  ds    E  dr rˆ   
dr
2
40 r
i


f
q 1 R
VR  V 
|
40 r
q
1
VR 
40 R
8
Inside


q
ˆ
Einside 
rr 
rrˆ
3
3 0
40 R
r 
r

q
Vr  VRin    E  ds   
rdr
3
40 R
R
R
Vr  VRin  
Vr  VRin  
Vr  VRin  
q
80 R 3
3
r
3
r 
q
80 R
q
80 R
r 2 |rR
2
2
 R2

q
80 R
9
Voltage=Outside+Inside
V  VR  Vr  VRin
q
1
q
q
2


r 
3
40 R 80 R
80 R
q
1
q
q
2
V

r 
3
40 R 80 R
80 R
3q
q
2
2
V
R

r
3
3
80 R
80 R
V
q
80 R
2
2
(
3
R

r
)
3
10
Part b) What is the potential difference between a point on the
surface and the sphere’s center?
V
q
80 R
at
r0
V
3q
80 R
2
2
(
3
R

r
)
3
 VCenter
V  VR  VCenter 
q
40 R

3q
80 R

q
80 R
11
Okay, that is if V=0 at infinity what if
V=0 at the center of the sphere?

Einside 
qr
rrˆ
3
40 R
r
Vr  V0   
0
40 R
r
Vr  V0  
0
at r  R
q
q
40 R
3
3
rdr
rdr  V0 
qr
VRin  
2
80 R
3
q
80 R
But V0  0
qr 2
Vr  
80 R 3
Same as previous
Same as previous
12
Problem
The electric potential at points in a space
are given by
V=2x2-3y2+5z3
What is the magnitude and direction of the
electric field at the point (3,2,-1)?
13
E=-grad(V)

E  V
V  2 x 2  3 y 2  5z 3
V
V
V
 4x
 6 y
 15 z 2
x
y
z

E (3,2,1)  4(3) xˆ  6(2) yˆ  15(1) 2 zˆ

E (3,2,1)  12 xˆ  12 yˆ  15 zˆ

E  12 2  12 2  152  22.6 N / C
14
Directions

E (3,2,1)  12 xˆ  12 yˆ  15 zˆ
y 12
tan   
 1
x  12
1
0
  tan (1)  135
Direction w.r.t +x axis
z
2
2
2
cos   where r  x  y  z  22.6
r
1  15
0
  cos
 138.4
Direction w.r.t +z axis
22.6
15
Problem
Three +0.12 C charges form an equilateral
triangle, 1.7 m on a side. Using energy
that is supplied at a rate of 0.83 kW, how
many days would be required to move
one of the charges to the midpoint of the
line joining the other two charges?
16
Draw It
Initially, this charge is
1.7 m from the other
two charges
0.12C
1.7 m
1.7 m
0.85 m
0.12C
1.7 m
0.85 m
0.12C
Finally, this charge is 0.85 m
from the other two charges
17
Potential Difference
V  V f  Vi
kq kq 2kq


L L
L
kq kq 4kq
Vf 


L
L
L
2
2
4kq 2kq
V  V f  Vi 

L
L
2kq
V 
L
where
Vi 
L  1.7 m q  0.12C
18
Potential Energy
2kq2
U  qV 
L
2(9 x109 )(0.12) 2
U 
 152 x106 J
1.7
where
L  1.7 m q  0.12C
19
Power=Work per unit time
P=W/t
 W=-U
 So 0.83 kW= 830 J/s
 And t= U/P=152x106/830
 t=183,699 s or 51 hours or 2.12 days

20