Gauss’s Law
22-17
(b)
(i) small shell inner surface: Since a Gaussian surface with radius
r
, for
arb
<
<
, must enclose zero net
charge, the charge on this surface is zero.
(ii)
small shell outer surface:
2
q
+
.
(iii)
large shell inner surface: Since
a Gaussian surface with radius
r
, for
crd
<
<
, must enclose zero net charge, the charge on this surface is
2
q
−
.
(iv)
large shell outer surface: Since there is
2
q
−
on the inner surface and the total charge on this conductor is
2
q
−
,
the charge on this surface is zero.
EVALUATE:
The outer shell has no effect on the electric field for
rc
<
. For
rd
>
the electric field is due only to
the charge on the outer surface of the larger shell.
Figure 22.46
22.47.
IDENTIFY:
Apply Gauss’s law
SET UP:
Use a Gaussian surface that is a sphere of radius
r
and that is concentric with the charged shells.
EXECUTE:
(a)
(i) For
,0
,
raE
<=
since charge enclosed is zero.
(ii)
,
arbE
<
since the points are
within the conducting material.
(iii) For
2
0
2
1
,,
4
q
brcE
π
r
<<
=
P
outward, since charge enclosed is +2
q
.
(iv) For
,
crdE
=
since the points are within the conducting material.
(v)
For
2
0
2
1
4
q
rdE
π
r
>=
P
inward,
since charge enclosed is –2
q
. The graph of the radial component of the electric field versus
r
is sketched in
Figure 22.47, where we use the convention that outward field is positive and inward field is negative.
(b)
(i)
small shell inner surface: Since a Gaussian surface with radius
r
, for
a
<
r
<
b
, must enclose zero net
charge, the charge on this surface is zero.
(ii)
small shell outer surface: +2
q
.
(iii)
large shell inner surface: Since
a Gaussian surface with radius
r
, for
c
<
r
<
d
, must enclose zero net charge, the charge on this surface is –2
q
.
(iv) large shell outer surface: Since there is –2
q
on the inner surface and the total charge on this conductor is –4
q
,
the charge on this surface is –2
q
.
EVALUATE:
The outer shell has no effect on the electric field for
<
. For
>
the electric field is due only to
the charge on the outer surface of the larger shell.
Figure 22.47
22.48.
IDENTIFY:
Apply Gauss’s law.
SET UP:
Use a Gaussian surface that is a sphere of radius
r
and that is concentric with the sphere and shell. The
volume of the insulating shell is
()
33
3
42
8
[2 ]
VR
RR
π
=−
=
.
EXECUTE:
(a)
Zero net charge requires that
3
28
3
πρ
R
Q
−=
, so
3
28
Q
ρ
π
R
3
.
(b)
For
,
0
rRE
since this region is within the conducting sphere. For
2,
0
,
rR
E
since the net charge enclosed
by the Gaussian surface with this radius is zero. For
2
R
<
<
, Gauss’s law gives
23
3
00
4
(4
)
(
)
3
Q
E
π
rr
R
=+
−
PP
and
22
43
Q
ρ
E
π
−
. Substituting
ρ
from part (a) gives
0
0
2
.
7
28
QQ
r
E
π
r
π
R
P
P
The net enclosed charge for
each
r
in this range is positive and the electric field is outward.