Chapter 22--Examples
1
Problem

A charge, +q, is surrounded by a thin,
spherical shell of radius, a, which has a
charge density of –s on its surface. This shell
is, in turn, surrounded by another thin shell of
radius, b, which has a surface charge of +s.
Find the electric fields in
+s



Region 1: r<a
Region 2: a<= r <=b
Region 3: r> b
-s
b
+q
a
2
Step 1: Pick your shape

I choose spherical!
 
 So
 E  dA  E (4r
2
)
3
Region 1: r<a

+s
qenclosed=q
-s
b
+q


E 4r 2 
E
qenclosed
0

q
a
0
1
q
40 r 2
(radially outward )

1 q
E
rˆ (r  a )
2
40 r
4
Region 2: a<=r<=b

+s
qenclosed=q+(4a2)*(-s)
-s
b
+q
E 4r  
2
E
qenclosed
0

1
0
1
q  4sa 
40
r2
q  4sa 
2
a
2
(radially outward )

1 q  4sa 2 
E
rˆ (a  r  b)
2
40
r
5
Region 3: r>b

+s
qenclosed=q+(4a2)*(-s)+
(4b2)*(s)
-s
b
+q
E 4r  
2
E
1
40
qenclosed
0

q  4sa
1
0
2
q  4sa
2
 4sb
2

a
 4sb 2 
r2
(radially outward )

1 q  4sa 2  4sb 2 
E
rˆ (a  r  b)
2
40
r
6
Problem
An electric filed given by E=4i-3(y2+2)j
pierces the Gaussian cube shown below.
(E is in newtons/coulomb and y is
meters). What net charge is enclosed by
the Gaussian cube?
y
x
z
X=1.0 m
X=3.0 m
7
First, let’s get a sense of direction
4
y
 Planes
Normal to +x
2.x-z: Normal to –y
3.y-z: Normal to –x
4.x-z: Normal to +y
5.x-y: Normal to +z
6.x-y: Normal to -z
6
1.y-z:
1
3
x
2
z
X=1.0 m
5
X=3.0 m
8
Need to find qenclosed
qenclosed
 
  E  dA
qenclosed
 
  0  E  dA
qenclosed
 
 
 
 
 
  
  0   E  dA   E  dA   E  dA   E  dA   E  dA   E  dA 
2
3
4
5
6
1

0
9
Integrating each side (start with
surface 1)
 

 E  dA  E  xˆdydz
1
1
but

2
ˆ
E  4 x  3 y  2 yˆ
  2 2
 E  dA  dz  4dy  4 * 2 * 2  16


0
1
0
Region 3, in which the normal vector points
in the opposite direction, will have a value of
-16
10
The rest of the sides
Since E is perpendicular to sides 5 & 6, the result is zero.
3

2
2
ˆ
 E   y dxdz   dz   3( y y 0  2)dx  6(2)(2)  24
0
2
1

2
2
 E   yˆ dxdz   dz   3( y y 2  2)dx  6(4  2)(2)  72
3
0
4
1
 
N  m2
 E  dA 16  24  16  72  0  0  48 C
12
10
qenclosed  48 0  48(8.85 10 )  4.2 10 C
11
Problem
The figure below shows a crosssection of two thin
concentric cylinders with
radii of a and b where b>a.
The cylinders equal and
opposite charges per unit
length of l.
a)
Prove that E = 0 for r>a
b)
Prove that E=0 for r>b
c)
Prove that, for a<r<b,
E
1
l
l
a
b
l
20 r
12
First, I choose a shape

I choose cylindrical!
 
 So
E

d
A

E
(
2

rL
)

13
For r<a

qenclosed =0
l
 
 E  dA  E (2rL)
E (2rL )  0
l
a
b
E0
14
For a<r<b

qenclosed =lL
l
 
 E  dA  E (2rL)
lL
E (2rL ) 
0
1 l
E
20 r
l
a
b
15
For r>b

qenclosed =lL-lL=0
l
 
 E  dA  E (2rL)
E (2rL )  0
l
a
b
E0
This is the principle of a coaxial cable
16
Problem
A very long, solid insulating cylinder
with radius R has a cylindrical hole
with radius, a, bored along its entire
length. The axis of the hole is a
distance b from the axis of the
cylinder, where a<b<R. The solid
material of the cylinder has a
uniform charge density, p.
Find the magnitude and direction of
the electric field inside the hole and
show that E is uniform over the
entire hole.
R
b
a
17
First, let’s do a solid cylinder of
radius, R
 
 E  dA  E (2rL)

qenclosed  V   r 2 L
 r 2 L 
E (2rL ) 
0
E

E
1
20
1
20

r
rrˆ 
1
20

r
18
Now what if we have an off-axis
cylinder

We learned in Phys 250, that
we can “translate”
coordinates by r’=r-b
 Where


b is the direction and
distance of the center of
the off-axis cylinder
r is a vector from the
origin
 
 E  dA  E (2r ' L)

qenclosed  V   r '2 L
 r '2 L 
E (2rL ) 
0
E

E
b
r
r’

E
1
20
1
20
1
20

r '
r ' rˆ 

 
 r b
1
20

r '

19
Ehole= Esolid cylinder-Eoff-axis hole

Esolid 

1
20

Eoff  axis 
r
1
20

 
 r b




Ehole  Esolid  Eoff  axis



 


1
1
1
1
1
Ehole 
r 
 r b 
r 
r 
b
20
20
20
20
20


1
Ehole 
b
20


All of these are constants and do not depend on r.
20