Physics 7B
Fall, 2006
Homework 7 Solutions
22.12
(
a
)
The charge on a conducting sphere must be on the surface. If we construct a spherical Gaussian
surface inside the metal sphere, there will be no enclosed charge and thus
E
a
= 0
.
(
b
) Because the point is still inside the metal sphere, we have
E
b
= 0
.
(
c
)
Because the point is outside the metal sphere, the field is radial, given by
E
c
=
Q
/4
π
Å
0
r
c
2
= (– 3.50
×
10
–6
C)/4
π
(8.85
×
10
–12
C
2
/N
·
m
2
)(3.10 m)
2
=
– 3.28
×
10
3
N/C (toward the sphere).
(
d
) Because the point is outside the metal sphere, the field is radial, given by
E
d
=
Q
/4
π
Å
0
r
d
2
= (– 3.50
×
10
–6
C)/4
π
(8.85
×
10
–12
C
2
/N
·
m
2
)(6.00 m)
2
=
– 8.75
×
10
2
N/C (toward the sphere).
(
e
)
Because the charge on the shell is the same as the charge on the surface of the metal sphere, all of
the fields will be
the same
.
(
f
)
For points inside the nonconducting sphere, only the charge inside a spherical surface with a
radius
to the point will provide the field:
E
(
r
≤
r
0
) = [(
Q
/
9
π
r
0
3
)(
9
π
r
3
)]/4
π
Å
0
r
2
=
Qr
/4
π
Å
0
r
0
3
.
Thus we have
E
a
= (– 3.50
×
10
–6
C)(0.15 m)/4
π
(8.85
×
10
–12
C
2
/N
·
m
2
)(3.00 m)
3
=
– 1.75
×
10
2
N/C (toward the center).
E
b
= (– 3.50
×
10
–6
C)(2.90 m)/4
π
(8.85
×
10
–12
C
2
/N
·
m
2
)(3.00 m)
3
=
– 3.38
×
10
3
N/C (toward the center).
At points outside the nonconducting sphere, the field will be the same as before;
E
c
= – 3.28
×
10
3
N/C (toward the sphere);
E
d
= – 8.75
×
10
2
N/C (toward the sphere).
22.22
[For reference for problem 22.23]
From the symmetry of the charge distribution, we know that the
electric field must be radial, with a magnitude independent of the
direction. For a Gaussian surface we choose a sphere of radius
r
.
On this surface, the field has a constant magnitude and
E
and d
A
are parallel, so we have
E ·
d
A
=
E
d
A
. The charge density is
ρ
=
Q
/[
9
π
(
r
0
3
–
r
1
3
)].
(
a
) For the region where
r
<
r
1
, there is no charge inside the
Gaussian surface, so we have
E
= 0;
r
<
r
1
.
(
b
) For the region where
r
1
<
r
<
r
0
, we apply Gauss’s law:
ı
E
·
d
A
=
EA
=
Q
enclosed
/
ε
0
;
r
r
1
r
0
Q
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E
4
π
r
2
=
ρ
9
π
(
r
3
–
r
1
3
)/
Å
0
, which gives
E
=
ρ
9
π
(
r
3
–
r
1
3
)/4
π
Å
0
r
2
=
Q
(
r
3
–
r
1
3
)/4
π
Å
0
(
r
0
3
–
r
1
3
)
r
2
;
r
1
<
r
<
r
0
.
(
c
) For the region where
r
>
r
0
, the electric field is that of a point charge;
E
=
Q
/4
π
Å
0
r
2
;
r
>
r
0
.
22.23
From the symmetry of the charge distribution, we know that the
electric field must be radial, with a magnitude independent of
the direction. The charge density of the sphere is
ρ
=
Q
/[
9
π
(
r
0
3
–
r
1
3
)].
We can add the field of the point charge at the center to the
fields found in Problem 22:
(
a
) For the region where
r
<
r
1
, the electric field is that of
the point charge at the center:
E
=
q
/4
π
Å
0
r
2
;
r
<
r
1
.
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 Fall '08
 Packard
 Charge, Electrostatics, Work, Electric charge, Fundamental physics concepts, gaussian surface

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