roofner (bar784) – Homework 5 – Weathers – (22202)
1
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printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
On a clear, sunny day, there is a vertical elec
trical field of about 105 N
/
C pointing down
over flat ground or water.
What is the magnitude of the surface charge
density on the ground for these conditions?
Correct answer: 9
.
2969
×
10
−
10
C
/
m
2
.
Explanation:
Let :
E
=

105 N
/
C
.
By Gauss’ law,
contintegraldisplay
vector
E
·
d
vector
A
=
q
ǫ
0
.
If we consider the electric field as being im
mediately above the ground or water, then we
can think of the ground or water as an infi
nite sheet of charge. The question is, should
the electric field beneath the surface be the
same as the field above?
Should it be zero?
The real answer is quite complicated. In or
der to work this problem, we assume the field
is zero beneath the surface. This makes sense:
since water is a conductor, the field inside the
water should be zero. Using Gauss’ Law,
E A
=
σ A
ǫ
0
,
we have
σ
=
E ǫ
0
= (

105 N
/
C)
×
(
8
.
85419
×
10
−
12
C
2
/
N
·
m
2
)
=

9
.
2969
×
10
−
10
C
/
m
2
,
with a magnitude of 9
.
2969
×
10
−
10
C
/
m
2
.
002
(part 1 of 2) 10.0 points
A conducting spherical shell having an inner
radius of 4
.
4 cm and outer radius of 5
.
1 cm
carries a net charge of 7
.
2
μ
C. A conducting
sphere is placed at the center of this shell
having a radius of 1
.
3 cm carring a net charge
of 2
.
4
μ
C.
1
.
3 cm
4
.
4 cm
,
Q
′
2
inside
5
.
1 cm
Q
′′
2
outside
2
.
4
μ
C
7
.
2
μ
C
P
Determine the surface charge density on the
inner surface of the shell.
Correct answer:

9
.
86498
×
10
−
5
C
/
m
2
.
Explanation:
Let :
r
= 1
.
3 cm
,
not required
a
= 4
.
4 cm = 0
.
044 m
,
b
= 5
.
1 cm = 0
.
051 m
,
and
q
= 2
.
4
μ
C = 2
.
4
×
10
−
6
C
.
Basic Concept:
E
= 0 inside a Conductor.
Gauss’ Law,
contintegraldisplay
vector
E
·
d
vector
A
=
Q
ǫ
0
.
Solution:
Since the electric field is zero
inside any conductor in electrostatic equi
librium, the net charge is zero inside any
spherical Gaussian surface of radius
r
, where
a < r < b
. Thus the charge on the inner sur
face of the sphere must be

q
. If we call the
charge density on the inner surface
σ
in
, then

q
= 4
π a
2
σ
in
σ
in
=

q
4
π a
2
=

2
.
4
×
10
−
6
C
4
π
(0
.
044 m)
2
=

9
.
86498
×
10
−
5
C
/
m
2
.
003
(part 2 of 2) 10.0 points
Determine the surface charge density on the
outer surface of the shell.
Correct answer: 0
.
000293712 C
/
m
2
.
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roofner (bar784) – Homework 5 – Weathers – (22202)
2
Explanation:
Let :
Q
= 7
.
2
μ
C = 7
.
2
×
10
−
6
C
.
The total charge contained in any spherical
Gaussian surface of radius
r > b
must be
Q
+
q
. Since the charge at the origin and the
charge on the inner surface of the sphere are
equal and opposite, the charge on the outer
surface must be
Q
+
q
.
We find
σ
out
in a
manner similar to that used in Part 1
Q
+
q
= 4
π b
2
σ
out
σ
out
=
Q
+
q
4
π b
2
=
7
.
2
×
10
−
6
C + 2
.
4
×
10
−
6
C
4
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 Fall '00
 Littler
 Electrostatics, Magnetism, Potential Energy, Work, Electric charge

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