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chapter 24
Gauss’s law
Problem
32. Repeat Problem 26, assuming that Fig. 24-45 represents the cross section of a long, thick-walled pipe. Now the case
a
=
0
should be consistent with the result of Problem 31 for the interior of the rod.
Solution
Suppose that the pipe is long enough that the line symmetric result in Equation 24-8 can be used in Gauss’s law. Then
E
=
q
enclosed
=
2
πε
0
r
l
.
For
a
<
r
<
b
,
q
enclosed
=
ρ
V
=
ρπ
(
r
2
−
a
2
)
l
,
so
E
(
r
)
=
(
=
2
ε
0
)(
r
−
a
2
=
r
).
For
a
→
0,
the field
inside a uniformly charged solid rod in Problem 31(b) is recaptured.
Problem
66. The volume charge density inside a solid sphere of radius
a
is given by
=
0
r
=
a
,
where
0
is a constant. Find (a) the
total charge and (b) the electric field strength within the sphere, as a function of distance
r
from the center.
(a) The charge inside a sphere of radius
r
≤
a
is
q
(
r
)
=
∫
0
r
dV
.
For volume elements, take concentric shells of radius
r
and thickness
dr
, so
dV
=
4
π
r
2
dr
.
Then
q
(
r
)
=
4
r
2
=
4
(
0
=
a
)
r
3
=
πρ
0
r
4
=
a
.
0
r
∫
0
r
∫
For
r
=
a
,
the total charge is
0
a
3
.
(b) For spherical symmetry, Gauss’s law and Equation 24-5 give
4
r
2
E
(
r
)
=
q
(
r
)
=
0
=
0
r
4
=
0
a
,
or
E
(
r
)
=
0
r
2
=
4
0
a
.
Problem
72. An infinitely long nonconducting rod of radius
R
carries a volume charge density given by
=
0
(
r
=
R
),
where
0
is a
constant. Find the electric field strength (a) inside and (b) outside the rod, as functions of the distance
r
from the rod
axis.
Line symmetry, Equation 24-8, and Gauss’s law give a field strength of
E
=
λ
enclosed
=
2
0
r
,
where
enclosed
=
∫
0
r
dV
is
the charge within a unit length of coaxial cylindrical surface of radius
r
, and
dV
=
2
r
is the volume element for a unit
length of thin shell with this surface. (a) For
r
<
R
(inside rod),
enclosed
=
∫
0
r
(2
0
=
R
)
r
2
=
2
0
r
3
=
3
R
,
hence
E
=
0
r
2
=
3
0
R
.
(b) For
r
>
R
(outside rod),
enclosed
=
∫
0
R
(2
0
=
R
)
r
2
=
2
0
R
2
=
3
, hence
E
=
0
R
2
=
3
0
r
.
CHAPTER 25
ELECTRIC POTENTIAL
Problem
6.
A charge of
3.1 C
moves from the positive to the negative terminal of a 9.0-V battery. How much energy does the
battery impart to the charge?
Δ
U
AB
=
q
Δ
V
=
(3.1 C)(9.0 V)
=
27.9 J.

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