Physics 2049 Exam 2 Solutions
Friday, October 12, 2001
1. Three point charges are located at the vertices of an equilateral triangle of side d as shown in the figure.
If d = 1 cm, q1 = q2 = −10 nC, and q3 = −20 nC, how much work was required to set up this charge
configuration (assuming that initially they were infinitely far apart)?
q1
d
q2
q3
Answer: 4.5 × 10−4 J
Solution: The work required to assemble all of the charges is
W =
kq1 q3
kq2 q3
kq1 q2
+
+
.
d
d
d
(1)
Substituting the numbers gives 4.5 × 10−4 J.
2. A certain arrangement of charges produces an electric potential V (x, y, z) = a/(x2 + y 2 + z 2 ), with
a = 5.0 V · m2 . What is the magnitude of the electric field at the point x = 0.0 m, y = 3.0 m,
z = 4.0 m?
Answer: 0.080 V/m
Solution: The components of the electric field are given by Ex = −∂V /∂x, Ey = −∂V /∂y, and
Ez = −∂V /∂z. Taking the derivatives, we obtain
Ex =
2ax
,
(x2 + y 2 + z 2 )2
Ex =
2ay
,
(x2 + y 2 + z 2 )2
The magnitude of the electric field is
q
~ = E2 + E2 + E2 =
|E|
x
y
z
Ex =
2az
.
(x2 + y 2 + z 2 )2
2a
.
(x2 + y 2 + z 2 )3/2
(2)
(3)
Substituting the numbers yields 0.080 V/m.
3. A charge Q = 27 × 10−9 C is distributed uniformly on a string which is stretched between x = 0 and
x = 3.0 m. What is the potential V (x = 4.0 m)? Assume that the potential is zero at infinity.
y
What’s the potential here?
0
3.0
4.0
x
Answer: 112 V
Solution: Let L be the length of the string, so that the charge per unit length is λ = Q/L. Consider
an element of charge dq = λ dx which is a distance x from the right end of the string, and therefore a
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distance x + 1 from the point x = 4.0 m; this makes a contribution dV = kλ dx/(x + 1) to the potential.
Integrating this to obtain the net potential, we have
Z 3
kλ dx
V =
0 x+1
= kλ ln(4)
(9.0 × 109 Nm2 /C2 )(27 × 10−9 C) ln(4)
=
3.0 m
= 112 V.
(4)
4. A spherical drop of liquid mercury of radius R has a capacitance C. If two such drops combine to form
a single larger drop, what is its capacitance?
Answer: 1.26 C
Solution: The capacitance of a conducting sphere of radius R is C = 4π0 R; i.e., it is proportional to
the radius. If we combine two droplets, the volume of the combined droplet is twice that of the smaller
droplets, and since the volume is proportional to the radius cubed, the radius of the combined droplet
is 21/3 = 1.26 times that of the smaller droplets, so the capacitance increases by the same factor.
5. Four capacitors, each with a capacitance of 10 µF, are connected as shown in the figure to a 10 V
battery. What is the potential difference across capacitor C1 ?
C1
+
10 V -
Answer: 4 V
Solution: This is essentially Problem 26–16 of HRW (one of the suggested homework problems). In
the upper network of three capacitors, the two in series have an equivalent capacitance of 5 µF; this
is in parallel with a 10 µF capacitor, for an equivalent capacitance of 15 µF for the upper network.
This is in series with the lower 10 µF capacitor, for a total equivalent capacitance of 6 µF. Therefore,
the charge q on the lower capacitor is q = Ceq V = 60 µC, and the potential drop across the lower
capacitor is 6 V. Therefore, the potential drop across the upper network is 4 V, which is the potential
drop across C1 .
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6. A cross section of a hollow conducting cylinder, with inner radius a and outer radius b, is shown below.
The cylinder carries a current out of the page, with the current density given by J = cr2 , with r the
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distance from the axis of the cylinder and c = 3.0 × 106 A/m . If a = 1.0 cm and b = 2.0 cm, what is
the total current carried by the conductor?
b
a
Answer: 0.71 A
Solution: The total current through the cylinder is
Z
~
J~ · dA
i =
Z
=
a
=
b
J(r)(2πr dr)
πc 4
(b − a4 ).
2
(5)
Substituting the numbers, we obtain 0.71 A.
7. A uniform piece of aluminum wire is cut into two pieces of equal length, which are then connected in
parallel. What is the equivalent resistance of this combination?
Answer: 0.25 R
Solution: Since the resistance of a piece of wire is proportional to its length, after cutting each piece
has a resistance of R/2. Connecting these in parallel gives an equivalent resistance of R/4.
8. You want to use a 12 V, 30 W light bulb (designed for use in a car) in your home. In order that the
bulb function properly (i.e., so that you don’t exceed the recommended 12 V), you need to connect it in
series with a resistor R before connecting it to the wall outlet. Assuming that the potential difference
across the wall outlet is 120 V d.c. (direct current), what should R be?
Answer: 43.2 Ω
Solution: The resistance of the lightbulb is Rbulb = V 2 /P = (12 V)2 /(30 W) = 4.8 Ω. When connecting
this to the outlet in series with the resistor we want 12 V across the lightbulb and 108 V across the
resistor; since the current is the same through both we have (12 V)/(4.8 Ω) = (108 V)/R, so we find
R = 43.2 Ω.
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9. In the circuit shown below, what is the potential difference across the 3 Ω resistor?
6.0 Ω
4.0 Ω
12.0 Ω
12.0 V
3.0 Ω
5.0 Ω
Answer: 4.5 V
Solution: The 6 Ω resistor and 12 Ω resistor in parallel are equivalent to a 4 Ω resistor, which is in
series with another 4 Ω resistor for an equivalent resistance of 8 Ω. This is in parallel with a series
combination of 3 Ω and 5 Ω which is equivalent to an 8 Ω resistor; i.e., we have two 8 Ω resistors in
parallel, for an equivalent resistance of 4 Ω. The battery emf is 12 V, so there is a current of 3.0 A
drawn from the battery. When this current encounters the junction it is split equally between the two
branches since they have the same equivalent resistance, so each branch has 1.5 A through it. The
potential drop across the 3 Ω resistor is therefore (1.5 A)(3 Ω) = 4.5 V.
10. In the circuit shown below the switch S is open and the capacitor is uncharged. We close the switch
and allow the capacitor to fully charge. If the switch is now opened, how much time will be required
for the potential difference across the capacitor to reach 6 V?
100 Ω
S
12 V
100 Ω
2.0 µF
Answer: 2.8 × 10−4 s
Solution: When the capacitor is fully charged it will have 12 V across it. Opening the switch removes
the battery from the circuit, and we have two 100 Ω resistors in series with a 2.0 µF capacitor. The
equivalent resistance of the circuit is 200 Ω, and the time constant τ = Req C = 4.0 × 10−4 s. The time
dependence of the potential difference across the capacitor plates is
V (t) = V0 e−t/τ .
(6)
t = τ ln[V0 /V (t)] = (4.0 × 10−4 s) ln 2 = 2.8 × 10−4 s.
(7)
Solving for the time t, we have
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