Physics 142!
Summer 2013
Exam I
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.!
All of the following statements about lines of the electrostatic field are true, but
which one follows from the field equation 
∫ E ⋅ dr = 0 ?
!
!
√ !
!
2.
Field lines do not cross each other.
Field lines start on positive charges and end on negative charges.
Field lines do not form closed curves.
Field lines do not penetrate a conductor.
Concerning the relationship between E and V, which of these is wrong?
!
√ !
If V is constant at all points of a solid object, then E = 0 within the object.
If E is not zero at a point, then V cannot be zero at that point.
!
As one moves away from a positive point charge both V and the
magnitude of E decrease.
!
If V depends only on the coordinate variables x and y then Ez = 0 .
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Physics 142!
3.!
Summer 2013
In Case 1 shown below we have a non-conducting sphere of radius R, centered at
the origin, with total charge –Q spread uniformly through its volume; inside the
sphere, at distance d from the origin, is a point charge +Q. In Case 2 we have a
dipole, with charges ±Q separated by distance d, and with the negative charge
located at the origin. The E-field in the two cases is the same:
!
Everywhere.
!
Nowhere.
√ !
!
Only at points at distance greater than R from the origin.
Only at points at distance less than R from the origin.
−Q
d
+Q
−Q
Case 1
d
+Q
Case 2
!
!
4.!
When two identical capacitors are connected in parallel across a battery the total
charge on the combination is Q. If they are discharged and then connected in
series across that battery the charge on each is:
!
!
Q/2.
√ !
Q/4.
!
2Q.
!
4Q.
[In parallel each has half the charge, so the capacitance of each is C = Q/ 2V . In
series the effective capacitance is C′ = 12 C = Q/ 4V , so the charge is C′V = Q/ 4 .]
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Physics 142!
Summer 2013
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.
√
6.!
T!
F
Lightning is more likely to strike a tall tree than a short one because the potential
difference between the thundercloud and the tall tree is larger.
T!
√
F
!
[The potential difference is the same. The E-field between the cloud and the tall
tree is larger because the distance is smaller.]
7.!
There is no charge on the walls of an empty cavity within a conductor, regardless
of what charges might be on the outer surface of the conductor.
√
T!
F
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Physics 142!
Summer 2013
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method
clearly. Problems carry the point values shown.
1.!
Four charges are fixed at the corners of a square of side a as
shown.
−q
+q
+q
−q
a.!
Find the E-field (magnitude and direction) at the
center of the square.
b.!
Find the potential at the center of the square. [Take
V(∞) = 0 .]
c.!
Find the total potential energy of the system.!
d.!
Suppose the positive charge at the top right corner is released from rest.
Describe its motion qualitatively.
[20 points]
a.!
Zero. The contributions of like charges at opposite corners cancel.
b.!
Zero. The contributions of opposite charges at adjacent corners cancel.
c.!
There are four pairs of opposite charges and two of like charges. We find
kq 2
kq 2
kq 2
U = −4
+2
=
(−4 + 2) .
a
a
2⋅a
d.!
Because the total energy is negative, the system is bound. The released charge will
move along the diagonal of the square toward the other positive charge, until it
reaches the point where the potential energy again reaches the value in (c). There
the kinetic energy is again zero, so it reverses direction and oscillates between this
point and the starting point.
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Physics 142!
2.!
Summer 2013
Two identical small charged non-conducting balls are threaded on a
non-conducting vertical pole as shown. The bottom ball rests on the
table; the top ball can move up or down the pole without friction.
The balls have mass m and charge Q. The vertical distance between
the centers of the balls is y.
a.!
y
Let y0 be the separation of the balls when the top one
remains at rest. Find y0 in terms of the given quantities
and constants.
b.!
Suppose the top ball is lifted to a height 2y0 and released from rest.
Explain (mostly with words) how you would calculate the height to which
it falls before coming momentarily to rest. [You are not asked to carry out
the calculation, only to describe the steps.]
c.!
!
a.!
Is the resulting oscillation of the top ball simple harmonic motion? How
do you know?
[15 points]
Set the gravitational force (down) equal to the Coulomb repulsion (up):
mg = kQ 2 / y02 . This gives y0 = kQ 2 /mg .
b.!
Let the total potential energy (gravitational plus electrostatic) at separation y be
U(y) . [With the usual choices for where potential energy is zero, we would have
U(y) = mgy + kQ 2 / y .] Then the ball will again be at rest at separation y1 , where
U(y1 ) = U(2y0 ) . [It turns out that y1 = 12 y0 .]
c.!
Since y1 cannot be zero the oscillation is not symmetric about the equilibrium
point, so it cannot be simple harmonic. Alternatives: U is not a quadratic function
of y; the net force is not a linear function of y.
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Physics 142!
3.!
!
Summer 2013
Two concentric conducting spherical shells are arranged as
shown. The inner shell has radius a and carries charge +Q.
Surrounding it is an uncharged non-conducting shell of
outer radius 2a, with dielectric constant κ = 2 . The outer
conducting shell has radius 3a and is grounded ( V = 0 ),
making the system a capacitor.
a.!
What is the E-field in the gap between the outer shell
and the non-conductor, as a function of r, the distance
from the center of the spheres?
b.!
What is the E-field within the non-conductor?
c.!
Find the potential at the inner conducting shell.
d.!
Find the capacitance of the system.
[20 points]
a.!
This is just the field of a point charge Q at the center, or E = kQ/r 2 .
b.!
This is the field in (a) reduced by 1/κ , so E′ = kQ/ 2r 2 .
c.!
We have V(a) = − ∫
a
1 ⎞ kQ ⎛ 1 1 ⎞
⎛ 1
Edr − ∫ E′ dr = kQ ⎜
− ⎟+
−
. Thus
3a
2a
⎝ 2a 3a ⎠
2 ⎜⎝ a 2a ⎟⎠
V(a) =
d.!
2a
kQ ⎛ 1 1 ⎞ 5kQ
+
=
.
a ⎜⎝ 6 4 ⎟⎠ 12a
We have C = Q/V(a) =
12a
.
5k
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Physics 142!
4.!
Summer 2013
A 1 µF parallel plate capacitor is fully charged by a 100 V battery.
a.!
What is the stored energy?!
b.!
With the battery remaining connected, a slab of dielectric with κ = 2 is
placed between the plates, filling the gap. Now what is the stored energy?
c.!
Now the battery is disconnected and the slab is removed. What is the final
potential difference?
d.!
Find the final stored energy and account for the difference between this
and your answer to (b). [Where does the energy come from or go?]
[20 points]
a.!
We have U0 = 12 C0V02 = 12 ⋅10−6 ⋅ (100)2 = 5 × 10−3 J.
b.!
Now U1 = 12 C1V02 = 12 ⋅ 2 × 10−6 ⋅ (100)2 = 10−2 J.
c.!
The new charge is Q = C1V0 = 2 × 10−6 ⋅100 = 2 × 10−4 . With the slab removed the
potential difference becomes V1 = Q/C0 = (2 × 10−4 )/10−6 = 200 V.
d.!
We have U 2 = Q 2 / 2C0 = (2 × 10−4 )2 /(2 × 10−6 ) = 2 × 10−2 J. The increased energy
comes from the work done by the agent who removed the slab.
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Physics 142!
Summer 2013
Exam I
25
Median = 67
St Dev = 13.7
Students
20
15
10
5
0
<40
40
50
8
60
70
80
90