1. technology and computing
2. electronic components

PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 1
Suppose a region of space has a uniform electric field directed
downwards as shown. Three points are labeled A, B, and C.
Points B and C are on the same horizontal line. Which statement
is correct
A
A. The voltage at all three locations is the same
B. The voltage at points B and C are equal, and the voltage
at point B is higher than the voltage at point A
B
C
C. The voltage at points B and C are equal and the voltage
at point B is lower than the voltage at point A
D. The voltage at point A is the highest, the voltage at point
C is the second highest, and the voltage at point B is the lowest
E. None of the above
Electric fields point “downhill” to lower electric potential
Question 2
You have two identical resistors. If you connect just one of them to a battery, there is 10 W of
heat dissipated in the resistor. If you connect both of them in parallel to the same battery, what
total heat will be dissipated?
A.
B.
C.
D.
E.
2W
5W
10 W
20 W
100 W
Connecting a second resistor in parallel doesn’t affect the first resistor, but dissipates the same
amount of power (since it is identical) thus the dissipated power doubles.
Page 1 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
The following three questions refer to this situation.
Two capacitors, labeled 1 and 2, are attached in
parallel to a battery of constant voltage V as shown.
Capacitor 1 has a charge Q (meaning +Q on one
plate, –Q on the other) and capacitor 2 has a charge
of 3Q.
V
C1 , Q
C2 , 3Q
Question 3
Which capacitor has the smallest capacitance?
A.
B.
C.
D.
capacitor 1
Smaller capacitors hold less charge for the same potential difference
capacitor 2
Both have the same capacitance.
impossible to tell from the information given
Question 4
Suppose the battery has variable voltage. If the voltage of the battery is doubled, how much
charge flows through the battery as its voltage is increased from V to 2V?
A.
B.
C.
D.
E.
zero charge
4Q
Doubling the potential difference doubles the charge, so another 4Q flows out
3Q
5Q
6Q
Question 5
By what factor does the energy stored in the two capacitors together increase, when the battery
voltage is doubled from V to 2V?
A.
B.
C.
D.
½ (stored energy halves)
one (no increase in energy stored)
two (stored energy doubles)
four (stored energy quadruples)
U  12 CV 2 , double V, quadruple U
E. None of these.
Page 2 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 6
In the circuit shown, the potential at B is defined to be 0 V. What is the potential at A?
A.
B.
C.
D.
E.
40 V
–40 V
50 V
80 V
20 V
Net 40 V clockwise over 40  gives 1 A current
clockwise. Walking clockwise from B we go up
60V across the battery, down 10V across the
resistor to reach A. So we went up 50 V net.
Question 7
The resistors and batteries in the circuit below are all identical. Which circuit(s) have the least
power delivered to it/them?
A.
B.
C.
D.
E.
Circuit 1
Circuit 2
Circuit 3
Circuit 1 = Circuit 2
Circuit 1 = Circuit 3
Parallel ideal batteries does nothing, series (3) doubles current
Question 8
An electron (charge q = –e) is released from rest at point in empty space some distance from a
carbon ion, which has charge q = +6e. The electron is initially at a point where the voltage due
to the positively charged carbon ion is 0.50 V. Some time after release, the electron is at point
where the voltage due to the carbon ion is 1.50V. At that later time, the kinetic energy of the
electron is..
A.
B.
C.
D.
E.
6.0 eV
1.6  10–19 eV
2.0 eV
1.0 eV
None of these.
The details of what is creating the potential is irrelevant. We have a change (loss) in potential
energy U  qV   e 1 V   1eV and hence we gained that much kinetic energy.
Page 3 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 9
The figure shows a spherical capacitor in which a conducting
sphere (labeled I) shares the same center as a conducting spherical
shell (labeled III). There is an air gap (labeled II) between the
conductors. The sphere has a total charge of +Q and the spherical
shell has a total charge of –Q. Which of the following statements
is FALSE?
A. The electric field in region IV (outside the capacitor) is 0.
TRUE – net enclosed charge is zero
B. The electric field within the metallic shell (region III) is 0. TRUE – inside conductor
C. The charge on the outer surface of the shell is 0. TRUE – all charge on inner surface
D. The electric field in region II (empty space between the two capacitor plates) is 0.
FALSE – there is a positive charge on the surface of the sphere, making E outward in II
E. The electric field in region I (within the core sphere) is 0. TRUE – inside conductor
Question 10
Two parallel-plate capacitors with different capacitance but the same plate separation are
connected in series to a battery. Both capacitors are filled with air. The quantity that is the same
for both capacitors when they are fully charged is:
A.
B.
C.
D.
E.
potential difference. FALSE – different Cs, in series so same Q, different Vs
stored energy. FALSE – different Cs, same Q, different U  Q 2 2C
energy density. FALSE – different electric fields, so different densities
electric field between the plates. FALSE – different areas, so different s
charge on the positive plate. TRUE – they are in series
Question 11
A uniform electric field E of magnitude 6,000 V/m exists in a region of space as shown. What is
the magnitude of the electric potential difference between points X and Y?
A.
B.
C.
D.
E.
0V
1800 V
2400 V
3000 V
4200 V
Walking from X to Y we only pick up a potential difference
walking along the field, so only the 0.4 m distance matters. We
thus are going downhill (from X to Y) by 2400 V
Page 4 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 12
A charged rod with total positive charge Q uniformly distributed along its length is positioned as
shown. The point A is located a distance r below the center of the rod. (It is a distance L from the
two ENDS of the rod) What can you conclude about the electric potential VA at point A?
A.
VA  kQ
r
B.
VA  kQ
r
C.
kQ
D.
VA  kQ
L
E.
VA  kQ
L
r
 VA  kQ
r
L
L
L
A
Some charge is further away than r (so pretending all of the charge is r away will give you a
maximum potential). Some charge is closer than L (so pretending all of the charge is L away
will give you a minimum potential). In reality the potential is somewhere in between.
Question 13
The capacitance of each of the four capacitors shown is 500 F. The voltmeter reads 1000 V.
The magnitude of the charge, in coulombs, on each capacitor plate is:
A.
B.
C.
D.
E.
0.2 C
0.25 C
0.5 C
20 C
50 C
The 1000 V potential is the potential across a single capacitor (that voltmeter is in parallel with a
single capacitor). So a single cap will have charge Q  CV  500 F  1000V  500mC
Page 5 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 14
Ariella connects two light bulbs in parallel across a real battery (i.e. the battery has some internal
resistance). She adds a third bulb to the circuit, also connecting it in parallel. What happens to
the current flowing through the original two bulbs?
A.
B.
C.
D.
E.
It stays the same.
It increases.
It decreases.
We need to know the value of the internal resistance to answer this.
We need to know more about the resistances of the bulbs to answer this.
When a real battery is asked to source more current, because of the internal resistance its
terminal voltage will decrease. Thus the previously connected lightbulbs will grow dimmer.
This is what happens if you try to start your car with the headlights on (in older cars anyway) –
the lights will temporarily go dim.
Question 15
Two pairs of hollow, spherical conducting shells are connected with wires and switches. The
system AB (where A and B are far apart) is extremely far from system CD. In both systems the
large shells have four times the radius of the small shells. Before the switches are closed each
pair has a charge of +20 nC on the small shell (A,C) and +60 nC on the large shell (B,D)
A
(+20 nC)
B
(+60 nC)
C
(+20 nC)
D
(+60 nC)
When the switches are closed, charge is free to flow along the conducting wires connecting the
spherical shells. Rank the electric charge on the shells A-D after the switches are closed.
A. QA = QB = QC = QD (the electric charge is the same for all four shells)
B. QC < QA < QB < QD
C. QA = QC < QB < QD
D. QC < QA = QB < QD
E. The ranking of the electric charge cannot be determined
All the charge on C flows out to D. Only part of A goes to B.
Page 6 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 16
Consider a thin, straight rod of length L located on the positive x-axis with its left end at x = 0.
A charge Q is uniformly distributed over the rod. Which of the expressions below gives the
magnitude of the electric potential at point P located at x = L + Y on a line that is collinear with
the rod?
L
A.
kQ
dx

L 0 L  x Y 
B.
kQ
dx

L 0  x  L 2
C.
kQ
dx

L 0  x  Y 2
D.
kQ
dx

L 0 x Y 
E.
kQ
dx

L 0 x  L Y 
L
L
L
L
You should draw a picture! The distance from a point at x on the rod to point P is given by
(L+Y)-x. So we have
L
dV 
k dq
k QL dx
kQ
dx

 V   dV 

L 0 L  x Y 
L  Y  x L  Y  x
Page 7 of 10
PHYS 212 – MT2
Spring 2013
Sample 2 Solutions
Question 17
The figure below shows the potential
due to two infinite sheets of charge
with charge per unit area 1 and 2 .
From examining this plot we can
deduce that
A. 1 and 2 have the same sign, and |1| > |2|
B. 1 and 2 have the same sign, and |1| = |2|
C. 1 and 2 have the same sign, and |1| < |2|
D. 1 and 2 have opposite signs, and |1| > |2|
E. 1 and 2 have opposite signs, and |1| = |2|
F. 1 and 2 have opposite signs, and |1| < |2|
Two applications of Gauss’s law will tell us all we need to know. But first we need to convert
the potential to electric field (slope). On the left the field is 10 V/m to the left. In between it is
2.5 V/m to the right. And on the right it is 10 V/m to the right. So draw a Gaussian pillbox with
ends at -2 m and 0m. The net flux is positive, (12.5 V/m)AGauss. For the right sheet draw a
Gaussian pillbox with ends at 0 m and 2 m. Now the flux is (7.5 V/m)AGauss. That is, sheet 2 has
more charge, because for it the electric field points outward on both sides while for sheet one the
field points in from the left and out to the right. Both sheets are positive though as they both
create net outward flux.
Page 8 of 10
PHYS 212 – MT2
Spring 2013
Question 18
The voltage between points a and b in the circuit
shown is measured with an ideal voltmeter. What
does the voltmeter read? (Hint: Recall that an
ideal voltmeter has infinite internal resistance.)
A.
B.
C.
D.
E.
Sample 2 Solutions
R
R
V
V/4
V/3
V/2
V
zero
a
R
b
Hooking a voltmeter between a & b means no current flows there, nor through the top right
resistor. So we just have a voltage divider (two in this case identical resistors in series across a
battery) and we are coming off the center of it, where the voltage is V/2.
Question 19
You build a spherically symmetric charge distribution using only charged and neutral conductors
(no insulators, no infinitesimally thin “charge sheets” or infinitely thick slabs; just solid and
hollow spherical conductors are allowed). You measure the following potential as a function of
radial distance from the center of your apparatus:
V
0
a
b
r
Based on the above information and plot, what can you say for certainty about the situation:
A. The central conductor is solid (i.e. there is conducting material at r = 0) Not necessarily
– it could also be hollow
B. There is non-zero net charge in the system NO – the field outside is zero so there must
be a zero net charge
C. There is zero charge density at r = b NO – this surface terminates the field lines in the
region between a & b, so there must be negative charges there.
D. You made a mistake in either measuring or plotting the potential NO – this is a perfectly
valid potential – net charge is zero, positive on some inner object radius r, negative on the
outer spherical shell, inner radius b
E. None of the above
Page 9 of 10
PHYS 212 – MT2
Spring 2013
Question 20
A parallel plate capacitor has a charge-potential difference relationship as
plotted in the solid line at right. It is connected to a battery and charged
so that its state is indicated by the big blue dot.
With the battery still connected, you do positive work on the plates of the
capacitor in order to modify its charge-potential difference relationship to
one of the two dashed lines. At which, if any, of the 4 labeled states does
the capacitor end up?
A.
B.
C.
D.
E.
Sample 2 Solutions
q
B
A
D
C
V
A
B
C
D
None of the above
If you are doing positive work on the plates, that means that you are doing something to them
that they don’t want to do. Since they have equal and opposite charges, they naturally want to
attract each other (though hooked to the battery this might not be energetically favorable for the
entire system). None-the-less, you must be pulling them apart in order to do positive work on
the plates.
The capacitor is hooked to a battery, so the potential drop won’t change. But the charge must.
As you pull them apart the potential would tend to rise (constant E, increasing d, V = Ed). But it
isn’t which must mean that the electric field, and hence the charge, are decreasing. So we are
going to point C.
Page 10 of 10