Name _________________
Solutions for Test 1
February 18, 2008
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
Possibly useful formulas:
Constants
e = 1.602 ×10−19 C
ke = 8.988 × 109 N ⋅ m 2 /C2
ε 0 = 8.854 ×10−12 C2 /N/m 2
ke = 1 4πε 0
Resistivity
E = Jρ
Resistance
R = ρL A
Charge on
Surface of
Conductor
E=
σ
nˆ
ε0
Parallel
Plate
Capacitor
Aκε 0
C=
d
Current Density
J = nqv d
J = ne2 Eτ m
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
Energy in a
Capacitor
2
U = 12 C ( ΔV )
Dipoles
τ = p×E
U = −p ⋅ E
u = 12 ε 0 E 2
Temperature Dependence
ρ = ρ 0 ⎡⎣1 + α (T − T0 ) ⎤⎦
R = R 0 ⎣⎡1 + α (T − T0 ) ⎦⎤
1. When you place a dipole in the presence of a constant electric field, what happens?
A) There is a net force on the dipole in the direction of the field
B) There is a net force on the dipole in the opposite direction of the field
C) There is a net force, but it depends on the direction the dipole is oriented
D) There is no net force, but there is a torque on the dipole
E) There is no net force, torque, or any other kind of force on a dipole
2. Which of the following makes capacitors useful in many devices?
A) They can store energy much longer than other devices, such as batteries or
gasoline
B) They can store energy and then release it very, very quickly
C) They obey the simple relationship ΔV = IR, which allows them to be used in many
circuits
D) They can store far more energy than other storage methods
E) They produce a lot of heat, which makes them good for producing lights, making
toasters, etc.
3. When you use the formula for the difference in electrical potential between two
points, ΔV = − ∫ E ⋅ ds , what path are you supposed to take between the two points
you are measuring the potential difference?
A) You must take a straight line
B) You must take a path that always runs parallel to the electric field
C) You must take a path that always runs perpendicular to the electric field
D) You must take a smooth circular path between the two points
E) It doesn’t matter what path you take; you get the same potential no matter
what
4. How does the electrical resistance of a superconductor compare with a really good
conventional conductor like silver?
A) It has no measurable resistance
B) It has significant resistance, but much less than silver
C) It has resistance comparable to silver
D) It has more resistance than silver, but not a lot more
E) It has much more resistance than silver
5. If you place charge on a solid conducting sphere, the charge will
A) All move to the surface of the sphere
B) Be spread throughout the sphere uniformly
C) Be concentrated at the center of the sphere
D) Go to either the center of the surface, depending on what other charges are nearby
E) Conductors cannot have net charge on them because their potential is always zero
6. Sketched at right is a Gaussian surface, with various
charges arranged as shown. Assume that any charge
that appears to be inside the surface really is inside it.
What is the total electric flux out of the surface?
A) + 2 μC/ε0
B) - 2 μC/ε0
C) + 4 μC/ε0
D) - 4 μC/ε0
E) +6 μC/ε0
+8 μC
+2 μC
-6 μC
-7 μC
+9 μC
7. How does the electric force between two protons 1 nm apart differ from the force
between two electrons 1 nm apart?
A) The electron force is smaller
B) The electron force is bigger
C) The electron force is the same size, but it is attractive, whereas the protons are
repulsive
D) The electron force is the same size, but it is repulsive, whereas the protons are
attractive
E) The forces are the exact same size, and both proton-proton and electronelectron forces are repulsive
8. Which of the following most accurately describes what happens to charges in an
insulator when electric forces are applied to it?
A) Charges are free to flow in it, so they shift large distances in reaction to the
electric field
B) A small fraction of the charge is free to flow, so this small fraction moves a long
distance
C) The charges cannot flow long distances, but they can move slightly, so they
shift their positions
D) Insulators don’t have charges that move, so there is no reaction at all
E) I have no idea; please mark this one wrong
9. Assume in the figure at right, the number of charges
+
–
–
drawn represents the relevant charge density of the
+
two types of charge carriers, whose charges are equal
–
+ –
+
and opposite, and the arrows represent the drift
+
+ –
velocities. What direction is the current density?
–
A) Right
B) Left
C) There is no current density
D) It is impossible to tell without knowing the drift velocity of each charge carrier
E) It is impossible to tell, even if you know the drift velocity of each charge carrier
10. The circuit symbol shown here is the symbol for a
A) Capacitor B) Battery
C) Ground D) Resistor E) Switch
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences) or
brief sketch (10 points each).
11. Explain qualitatively how a lightning rod works.
A lightning rod consists of a sharp conductor that is put on top of the object to
protect, and the other end is grounded. When there is a lot of electrical potential
difference between the ground and the atmosphere, a very large electric field is produced
at the tip, which causes coronal discharge which then drains away excess charge. This
makes it less likely the object will be struck by lightning.
12. What factors are important when selecting a dielectric material to put in a
capacitor?
The most important factors are that you choose a dielectric material with a large
dielectric constant (to increase the capacitance), and a high dielectric strength (so it
doesn’t catastrophically discharge). It is also a good idea to have a very good insulator
(so it doesn’t let charge leak away) and be able to make it very thin.
13. Copy the three charges sketched below onto your answer sheet. Then make a
qualitative sketch of the electric field lines. The numbers represent the relative
size of the different charges
The number of field lines coming out of each charge must be proportional to the
charge, and the lines must start from the positive charges and end on the negative
charges. Since there is more total positive charge than negative charge, there will be
some electric field lines running off to infinity from the positive charge.
-3
+8
-3
Part III: Calculation: [60 points] Choose three of the following four questions and
perform the indicated calculations (20 points each)
12 V
14. A 12.0 V battery is connected to three capacitors
as shown at right. C1 = 30.0 μF and C2 = 10.0
C1
μF. The third capacitor, C3, is a parallel plate
2
30.0 μF
capacitor with area A = 0.900 m , separation d =
1.00 μm, and initially nothing between the plates
(a) [5] What is the capacitance of the capacitor C3?
C3
C2
10.0 μF
The capacitance is given by the formula given for a parallel plate capacitor, but in
this case, since there is no dielectric, we use κ = 1, so we have
2
−12 2
2
2
Aκε 0 ( 0.900 m ) (1) ( 8.854 × 10 C /N/m )
−6 C
C=
=
=
7.969
×
10
d
1.00 ×10−6 m
N⋅m
C
= 7.969 × 10−6 = 7.969 μ F
V
(b) [8] What is the capacitance of the whole combination of the three capacitors?
The capacitors C2 and C3 are connected at both ends, and hence are connected in
parallel. The combined capacitance of this combination is, therefore,
C23 = C2 + C3 = 10.0 μ F + 7.97 μ F = 18.0 μ F
This combination can then be replaced by a single capacitor, and this is connected in
series with C1, since it is joined at the bottom and there is nothing else connected there.
The combination, therefore, has total capacitance
1
1
1
1
1
=
+ =
+
= 0.0890 μ F−1 ,
C C23 C1 18.0 μ F 30.0 μ F
C = 11.2 μ F
(c) [4] What is the total energy stored in the capacitors by the battery?
The energy stored in the capacitor is given by
U = 12 C ( ΔV ) =
2
1
2
(11.2 ×10
−6
F ) (12.0 V ) = 8.09 ×10−4 J = 0.809 mJ
2
(d) [3] What would be the capacitance C3 if you filled up the space with a
dielectric fluid with dielectric constant κ = 2.73?
The only change to part (a) is that the factor of κ would increase the capacitance
to C = κ C0 = 2.73 ( 7.969 μ F ) = 21.7 μ F
15. A wire is in the form of a long
Resistivities and temperature coefficients at 20° C
cylinder of radius r = 1.00 mm.
Temperature
In total the cylinder is 20.0 m
Material
Resistivity
(Ωm)
Coefficient
(°C-1)
long, with the first 15.0 m
Iron
10 × 10-8
5.0 × 10-3
consisting of iron and the last
-8
Lead
22 × 10
3.9 × 10-3
5.0 m consisting of lead.
(a) [9] What is the total resistance of the wire at 20.0° C?
Resistance can be calculated from the formula R = ρ L A , where A is the crosssectional area for a cylinder, which is a circle. The area is given by A = π r 2 . For the
resistivity, we can use the values given in the table above, since these values are at a
temperature of 20.0° C. Hence the resistance is
RIron
(10 ×10 Ωm ) (15.0 m ) = 0.477 Ω,
=
=
A
π (1.00 ×10 m )
ρ L ( 22 ×10 Ωm ) ( 5.0 m )
=
=
= 0.350 Ω.
A
π (1.00 ×10 m )
ρL
−8
−3
2
−3
2
−8
RLead
Now, if you connect two resistors end to end, the current through each of them must be
the same, and the total voltage change must be the sum of the voltages. As a
consequence, it is easy to show that the resistances add, so the total resistance is
R = RIron + RLead = 0.477 Ω + 0.350 Ω = 0.827 Ω
(b) [5] If the wire is connected to a 120 V power source, calculate the current I
and the power P consumed in the wire.
The current can be found from
I = ΔV R = 120 V 0.827 Ω = 145 A
The power can then be found from
P = I ΔV = (145 A )(120 V ) = 17.4 kW
(c) [6] Due to the heat in the wire, its temperature rises to 120.0° C. What is its
total resistance now?
As the temperature rises, each term in the resistance rises by a factor of
1 + α (T − T0 ) , so the new resistance for the iron and lead is
RIron = R Iron,0 ⎡⎣1 + α (T − T0 ) ⎤⎦ = ( 0.477 Ω ) ⎡⎣1 + .0050 (120 − 20 ) ⎤⎦ = 0.716 Ω
RIron = R Lead,0 ⎡⎣1 + α (T − T0 ) ⎤⎦ = ( 0.350 Ω ) ⎡⎣1 + .0039 (120 − 20 ) ⎤⎦ = 0.486 Ω
The total resistance, then, is 1.202 Ω.
16. A point positive charge q is placed at the
center of a hollow spherical conductor of
inner radius a and outer radius b. The
conductor itself has an additional negative
charge – 2q placed on it. Find the magnitude
and direction of the electric field at all points;
that is, for
(a) r < a
(b) a < r < b
(c) r > b [16 for all parts]
-2q
q
b
a
The problem is obviously spherically
symmetric, so it is well set-up for using Gauss’s
Law. If we draw a sphere of radius r around the center of the structure, the electric field
will certainly point radially outward and be the same in all directions, E = Erˆ . The flux
passing through a spherical surface at this radius will therefore be E times the area of a
sphere, so Φ E = 4π r 2 E . Gauss’s Law then tells us that
4π r 2 E = Φ E = qin ε 0 .
Then we solve for the electric field E and substitute back in to find
E = Erˆ =
kq
qin rˆ
= e 2in rˆ ,
2
4πε 0 r
r
where we have rewritten it in a slightly simpler form in terms of Coulomb’s constant. It
remains to find qin in each of the regions.
For r < a, qin = q, because that’s the only charge in that region. For a < r < b, we
are inside the conductor, so E = 0 automatically. And for b < r, the total charge inside is
the sum of the charges q and – 2q. So in summary, the electric field is
⎧ ke q ˆ
for r < a,
⎪ r2 r
⎪
for a < r < b, ,
E=⎨ 0
⎪ kq
⎪− e2 rˆ
for r > b.
⎩ r
The electric field is radially outward in the interior, has no direction inside the conductor,
and is inwards outside.
(d) [4] Find the total charge on the inner surface of the conductor.
Gauss’s law applies inside the conductor as well, which tells you the total charge
inside this region is 0. Since there is a charge q at the center, there must be a
compensating charge – q on the inside surface of the conductor. There must also be a
total charge – q on the outside of the conductor, but this wasn’t asked for.
17. As calculated in class, the electric potential a
distance z above the center of a disk of radius a
and charge density σ is given by:
V ( z ) = 2π keσ
(
z2 + a2 − z
P
z
)
a
(a) [8] What is the direction of the electric field at
the point P? Calculate a formula for Ez, the electric field in the z-direction at
P.
From the geometry of the problem, the disk will produce an electric field away
from the disk, which must be in the +z direction. This is fortunate, since we only know
the potential along the z-axis, and hence can’t take derivatives in the other two directions.
To find Ez, we take
Ez = −
∂V
∂
= −2π keσ
∂z
∂z
(
)
⎛
⎞
z
z 2 + a 2 − z = 2π keσ ⎜1 −
⎟
z2 + a2 ⎠
⎝
(b) [12] A small object with charge q = 2.70 μC and mass m = 5.10 g is placed at
rest at the point P. The disk has surface charge density σ = 470 μC/m2,
radius a = 1.1 m, and the small charge starts a distance z = 0.61 m away.The
small object is allowed to accelerate to infinity without encountering any
friction. Find its speed at infinity.
To find its speed at infinity, we simply use conservation of energy. The initial
potential energy of the charge is
U = qV = 2π keσ q
(
z2 + a2 − z
)
= 2π ( 8.988 ×109 N ⋅ m 2 / C2 )( 470 ×10−6 C/m 2 )( 2.70 ×10−6 C )
(
)
0.612 + 1.12 − 0.61 m
= 46.4 N ⋅ m = 46.4 J
As the object accelerates to infinity, the potential energy drops to zero, so there is a net
decrease of 46.4 J in its potential energy. This must be exactly offset by a 46.4 J increase
in its kinetic energy, given by
T = −ΔU = 12 mv 2
Solving for the final velocity v, we have
v2 =
2T 2 ( 46.4 J )
=
= 1.82 ×104 m 2 /s 2 ,
m 0.0051 kg
so
v = 135 m/s .