Physics 54 !
Summer 2012
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.
!
A small dipole with moment p oriented as shown is near a fixed positive point charge Q . If the dipole is released from rest in this configuration:
!
p will rotate clockwise and then be repelled by Q .
Q
!
p will rotate counter-clockwise and then be repelled by Q .
√
!
p will rotate clockwise and then be attracted by Q .
!
p will rotate counter-clockwise then be attracted by Q .
p
!
2.
!
Two hollow spheres have the same dimensions and carry the same total charge + Q . Sphere A is a conductor; sphere B is a non-conductor with charge distributed uniformly throughout the material. Consider each case separately as an isolated sphere, taking
V
( ∞ ) = 0 . The potential at the center of the sphere:
A B
!
Is the same for both cases.
!
Is equal to the potential at the outer surface for both cases.
!
Is greater for case A .
√
!
Is greater for case B .
[The potential continues to increase as we cross the charged material in the nonconducting sphere, while for the conductor the potential remains constant inside the outer surface.]
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Physics 54 !
Summer 2012
3.
!
Two point charges of equal magnitude are fixed at points on the x -axis. There is a point on that axis between the charges where V
= 0 (taking as usual
V
( ∞ ) = 0 ).
!
There is a point on the finite x -axis where E
= 0 .
!
There is another point on the finite x -axis where V
= 0 .
√
!
The potential energy of the charges is negative. {Opposite sign charges.]
!
None of the above is true.
!
!
4.
!
In the situation shown, a parallel plate capacitor is fully charged and remains connected to the battery. A conducting plate is inserted into the gap between the plates, partially filling the gap. Which of the following is NOT true?
!
The capacitance increases.
!
!
It makes no difference where the conducting plate is located vertically between the plates.
√
!
The E-field in the gaps between the capacitor plates is unchanged.
[It increases because the charge increases.]
!
The charge on the capacitor plates increases.
!
!
Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each question carries a value of 3 points.
5.
!
The total charge in a closed system cannot be changed by any means.
√
T F
6.
!
If a point charge Q is surrounded by a hollow metal container, the total charge on the inner surface of the container is – Q regardless of the shape of the container.
√
T F
7.
!
Removing dielectric material from between the plates of a charged and isolated capacitor results in an increase in the energy stored by the capacitor.
√
T F [ C decreases so U increases.]
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Physics 54 !
Summer 2012
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method clearly. Problems carry the point values shown.
1.
!
Two dipoles, consisting of charges ± q separated by distance d , oriented as shown, are a distance x apart.
x a.
!
Write the potential energy in terms of the given quantities (omitting the energy of the charges within each dipole).
b.
!
Use the binomial approximation to find the formula for this energy when
x
>> d . Write it in terms of the dipole moment
p
= qd .
c.
!
If the dipoles are released from rest, will they attract or repel each other?
How do you know?
[15 points] a.
!
The opposite charge pairs are at distance x apart. The same charge pairs are at distance
x
2 + d
2
apart. Thus we have
U = − 2 k q
2 x
+ 2 k x
2 q
2
+ d
2
= − 2 kq
2
⎛
⎝
1 x
− x
2
1
+ d
2
⎞
⎠
.
b.
!
The 2nd term in ( ) can be written as
1 x find U ≈ − k
( qd )
2 x
3
= − k p x
2
3
.
(
⋅ 1 + ( d / x )
2
) − 1/2
≈
1 x
(
1 − 1
2
( d / x )
2
)
, so we c.
!
Since U is negative the system is bound. They will attract each other.
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!
Physics 54 !
Summer 2012
2.
!
A parallel plate capacitor has half the region between its plates filled with a dielectric material as shown. The plates have area A and are separated by distance d . The potential of the positive plate is
V 0
and that of the negative plate is zero. Take κ = 2 .
+ a.
!
Let the E-field in the empty gap be
E 1
and that in the dielectric
E 2
. What is the ratio
E 2
/ E
1
?
x b.
!
What are
E 1
and
E 2
in terms of the given quantities?
[Express
V 0
in terms of the E-fields.] c.
!
Plot carefully the potential between the plates V vs. x , with x
= 0 at the positive plate. [What is the potential at
x
= d /2 ?]
− d.
!
Find the stored energy. [What are the energy densities in the two regions?]
[20 points] a.
!
E 2
/ E
1
= 1/ κ = 1/2 .
b.
!
We have
V 0
= E
1
⋅ ( d /2) + E
2
( d /2) . Using (a) we find E
1
=
4 V
0
3 d
and E
2
=
2 V
0
3 d
.
c.
!
See plot. The potential at
x
= d /2 is
2
3
V
0
. d.
!
We have u e
(1) = 1
2
ε
0
E
2
1 in each region is
A
= 8
9
ε
0
( V
0
/ d )
2
and u e
(2) = 1
2
κε
0
E
2
2
⋅ ( d /2) so we find U = ( Ad /2) ⋅ [ u e
=
(1) + u
4
9
ε
0
( V
0
/ d )
2
. The volume e
(2)] = 2
3
ε
0
A d
⋅ V
0
2
.
!
V 0
V
2
3
V
0
!
d
/2
4
d
x

Physics 54 !
Summer 2012
3.
!
In the cases shown, a conducting sphere of radius a is surrounded by a larger concentric conducting shell of radius b and negligible thickness.
Case 1 Case 2
Case 1. Charge + Q is placed on the inner sphere only.
Case 2. Charge +Q is placed on the outer shell only.
a.
In each case, sketch lines of the E-field, in the gap between the conductors and in the space outside the shell.
b.
!
In each case, find the potential on both conductors (taking
V
( ∞ ) = 0 ).
c.
!
Compare the energy stored in the field in the two cases. [Which is larger?] d.
!
Suppose a thin conducting wire is connected between the sphere and the shell. What difference would that make to your answers?
[20 points] a.
!
See drawing above. In Case 2 the charge is only on the outer surface of the shell.
b.
!
For r
V
> b the situation is like a point charge Q at the center in both cases, so
( b ) = kQ / b . In Case 2 the potential is constant at this value for r
< b . For Case 1 we have V ( a ) − V ( b ) = −
b a kQ r
2 dr = kQ
⎛
⎝⎜
1 a
−
1 b
⎞
⎠⎟
. Since V ( b ) = kQ b
we find V ( a ) = kQ a
.
[The negligible thickness of the shell makes it this simple.] c.
!
The energy stored outside the outer shell is the same in both cases. In Case 2 that is the total energy stored. But in Case 1 there is also energy between the sphere and the shell, so the total energy is greater in that case.
d.
!
No difference in Case 2, because the two conductors are already at the same potential. In Case 1 the charge on the inner sphere would cancel with the charge on the inner surface of the shell, making this case like Case 2.
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Physics 54 !
Summer 2012
4.
!
A parallel plate capacitor has plates of area A separated by distance x , so it has capacitance
C
= ε
0
A / x . It is charged to potential V by a battery, which remains connected to it. An external agent changes the distance between the plates from initial value
x 1
to final value
x 2
, changing the capacitance from
C 1 to
C 2
, and changing the stored energy by the amount Δ U = 1
2
( C
2
− C
1
) ⋅ V
2
.
a.
!
When the distance between the plates is x , what is the charge
Q
( x ) ?
b.
!
The force of attraction between the plates is the charge on one plate times the E-field created by the other. [Each plate creates half the total field in the gap.] Write this force in terms of
Q
( x ) and then use your answer to (a) to put it in terms of x .
c.
!
Show that the work done by the external agent is equal to −Δ U .
d.
!
Find the initial and final charges, and show that the work done by the battery while the charge changes is equal to 2 Δ U .
!
[20 points] a. !
Q ( x ) = C ( x ) ⋅ V =
ε
0
A x
⋅ V .
b.
!
The field due to one plate alone is E
= σ
/ 2
ε
0
=
Q / 2 A
ε
0
. The force is thus
F
= QE = Q
2
/2 A ε
0
. Substituting from (a) we have F ( x ) =
ε
0
AV
2
2
⋅
1 x
2
.
c.
!
W =
x
2 x
1
F ( x ) dx =
ε
0
AV
2
2
⋅
⎛
⎝⎜
1 x
1
−
1
⎞ x
2
⎠⎟
= 1
2
( C
1
− C
2
) ⋅ V
2
. This is −Δ U as claimed.
d.
!
The initial and final charges are supplied by the battery is
Q 1
Δ Q = ( C
2
= C
1
V and
Q 2
− C
1
) V
= C
2
V , so the amount of charge
. the work done to put this charge on the capacitor is
W batt
= Δ Q ⋅ V = ( C
2
− C
1
) ⋅ V
2 = 2 Δ U , as claimed.
6

Physics 54 !
Median = 63
SD = 11
Summer 2012
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