Phys 106 Final Examination
Page 1
Monday, 30-May, 2011
PROBLEM (1)
(20 points)
A solid insulating sphere of radius
R
has a volume charge
density that varies with
r
according to the expression
"
=
#
r
2
where
"
is a positive constant and
r
<
R
is
measured from the center of the sphere. Concentric with the
sphere is a conducting spherical shell with inner radius 2
R
and outer radius 3
R
, and the shell has a net positive charge
Q
.
Express your answers in terms of the given quantities and
fundamental constants as needed.
R
2
R
3
R
A
B
(a) (4 pts) Find the charge contained within the solid insulating sphere of radius
R
.
Consider a charge element
dq
as a spherical shell of radius
r
and thickness
dr
inside the insulating sphere. The spherical
shell has a volume
dV
=
4
"
r
2
dr
.
q
= charge contained within solid insulating sphere
q
=
dq
0
R
"
=
#
dV
0
R
"
=
$
r
2
(4
%
r
2
dr
)
0
R
"
q
=
4
"#
r
4
dr
0
R
$
=
4
"#
r
5
5
%
&
'
(
)
*
0
R
q
=
4
"#
R
5
5
(b) (6 pts) Using Gauss's law, find an expression for the electric field in the regions (i)
R
<
r
< 2
R
, (ii) 2
R
<
r
< 3
R
, and
(iii)
r
> 3
R
. State the direction of the electric field in each region.
(i) Consider a spherical gaussian surface of radius
r
such that
R
<
r
< 2
R
, and concentric with the solid sphere.
"
E
=
!
E
#
$
d
!
A
=
E
cos
%
dA
=
q
in
&
0
#
E
4
"
r
2
=
q
in
#
0
=
q
#
0
=
4
"$
R
5
5
#
0
E
=
"
R
5
5
#
0
r
2
(for
R
<
r
< 2
R
, directed radially outward)
(ii)
E
= 0 (for 2
R
<
r
< 3
R
, inside a conductor)
(iii) Following the steps in part (i):
E
4
"
r
2
=
q
in
#
0
=
q
+
Q
#
0
=
4
"$
R
5
5
#
0
+
Q
#
0
E
=
"
R
5
5
#
0
r
2
+
Q
4
$#
0
r
2
(for
r
> 3
R
., directed radially outward)
(c) (4 pts) Find the surface charge densities
!
in
and
!
out
on the inner and outer surfaces of the conducting spherical shell,
respectively.
Consider a spherical gaussian surface of radius
r
such that
R
<
r
< 3
R
, and concentric with the solid sphere.
"
E
=
!
E
#
$
d
!
A
=
E
cos
%
dA
=
q
in
&
0
#
E
= 0 and hence
q
in
= 0. So,
q
in
=
q
+
4
"
(2
R
)
2
#
in
=
0
"
in
=
#
q
16
$
R
2
=
#
1
16
$
R
2
4
$%
R
5
5
--->
"
in
=
#
$
R
3
20
Q
=
4
"
(2
R
)
2
#
in
+
4
"
(3
R
)
2
#
out
=
16
"
R
2
#
in
+
36
"
R
2
#
out
"
out
=
Q
36
#
R
2
$
16
#
R
2
"
in
36
#
R
2
=
Q
36
#
R
2
$
4
"
in
9
"
out
=
Q
36
#
R
2
$
4
9
(
$
%
R
3
20
)
--->
"
out
=
Q
36
#
R
2
+
$
R
3
45
Notice: There may be alternative solutions.
(d) (6 pts) Find the potential difference
V
B
!
V
A
between the points
A
and
B
, shown in the figure.
V
B
"
V
A
=
"
!
E
#
d
!
s
=
A
B
$
"
E
R
3
R
$
ds
(cos 0
°
)
=
"
Edr
R
3
R
$
V
B
"
V
A
=
"
Edr
R
2
R
#
"
Edr
2
R
3
R
#
=
"
Edr
R
2
R
#
"
0
V
B
"
V
A
=
"
#
R
5
5
$
0
r
2
dr
R
2
R
%
=
"
#
R
5
5
$
0
[
"
1
r
]
R
2
R
=
#
R
5
5
$
0
(
1
2
R
"
1
R
)
V
B
"
V
A
=
"
#
R
4
10
$
0

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