Name _________________
Solutions to Test 1 - February 10, 2012
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
Charge on
Parallel
Energy in a
Possibly useful formulas:
Surface of
Plate
Capacitor
2
Conductor
Capacitor
U  12 C  V 
Constants
A 0

e  1.602 1019 C
E  nˆ
C
u  12  0 E 2
9
2
2
d

0
ke  8.988  10 N  m /C
 0  8.854 1012 C2 /N/m 2
ke  1 4 0
Cones
V  13  R 2 L
Alat   R L2  R 2
Cylinders
V   R2 L
Alat  2 RL
Spheres
A  4 R 2
V  R
4
3
Dipoles
τ  pE
U  p  E
Triangles
A  12 BH
Circles
C  2 R
3
A   R2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. If you place some charge on a solid conducting sphere, the charge will move to
A) The center
B) The surface
C) It will be uniformly distributed throughout
D) Wherever you put it, since charges don’t move in a conductor
E) California
2. How do the charge of the proton and electron compare?
A) The magnitude of the proton’s charge is greater
B) The magnitude of the electron’s charge is greater
C) They are equal in magnitude, and both are positively charged
D) They are equal in magnitude, and both are negatively charged
E) They are equal in magnitude, but opposite in sign.
n
4. In the sketch at right, the electric field is in the +x direction, and
the rectangle drawn is in the xy-plane with a unit normal in the
+z direction. The total electric flux across the rectangle will be
A) Positive B) Negative C) Zero D) Insufficient information E) I have no idea
4. What is the significance of the circuit symbol sketched at right?
A) Anything attached here has potential zero
B) Anything attached here has no charge
C) Anything attached here is a conductor
D) This is a switch, meaning it can either be connected or disconnected
E) This is a capacitor, which stores charge
5. The unit of electric potential is
E
A) C
B) N
C) V/m
D) V
E) F
6. What effect does putting some sort of material between the two plates of a parallel
plate capacitor generally have on the capacitance?
A) It increases it
B) It decreases it
C) It leaves it unchanged
D) It can either increase or decrease it, depending on the material
E) It decreases it or leaves it the same, depending on the material
7. If you are just outside the surface of a conductor, which direction will the electric
field point?
A) It will always point outwards
B) It will always point inwards
C) It will point away from the conductor if the surface charge is positive, and
toward it if the surface charge is negative
D) It will point toward the conductor if the surface charge is positive, and away from
it if the surface charge is negative
E) It will point parallel to the surface of the conductor
8. Suppose we have two equal and opposite point charges +Q and –Q. Which of the
following is true of the point exactly half way between them?
A) The electric field and electric potential will both be zero
+Q
B) The electric field and electric potential will both not be zero.
C) The electric field will be zero, but the potential will not
D) The electric potential will be zero, but the electric field will not
E) There is insufficient information to know for sure.
9. A charge Q is placed in a spherical region of unknown radius, but not necessarily at
the center. What additional information is needed to determine what the total electric
flux out of the region is?
A) You need to know the radius of the sphere, and the exact position of the charge
B) You need to know the radius of the sphere, but not the position of the charge
C) You need to know the distance of the charge from the center of the sphere
D) You need to know the ratio of the distance of the charge to the radius
E) You don’t need any additional information
10. Which of the following is not true about how materials become charged?
A) Most ordinary objects contain a very large number of positive charges
B) Most ordinary objects contain a very large number of negative charges
C) An object could become positively charged by removing electrons from it
D) An object could become negatively charged by adding electrons from it
E) When an object becomes charged, most of the electrons in it have been
removed or added
Part II: Short answer [20 points]
–Q
Choose two of the following questions and give a short answer (2-3 sentences) or
brief sketch (10 points each).
11. Sketched at right are the electric field
lines from a pair of point charges A and
B. What is the sign of each charge,
which charge is larger, and how much
bigger (approximately) is the larger
charge?
A
B
Electric fields go from positive to negative
charges, so A must be positive and B
negative. The number of field lines is
proportional to the charge, so since there are
twice as many coming from A as from B, the
charge A must be about twice as big as B in magnitude.
12. Using the same diagram as for question 11, sketch in the approximate
equipotential lines. Include 3 or more curves, and indicate which ones represent
the highest potential and which the lowest potential.
I’ve sort of crudely sketched them in. They have to always be perpendicular to
the electric field lines, roughly. The highest ones are around the charge A and the lowest
ones around charge B. Don’t take my sketches too literally.
13. Explain qualitatively what an electric dipole is. If a dipole is placed in a constant
electric field, what will the total force on the dipole be? Will there be any sort of
other forces (not net force) on it?
An electric dipole is anything with total charge zero, but with more charge on one
side than the other. In the presence of a constant electric field, a dipole will feel no net
force, but the differences of the locations of the forces produces a torque, or twisting
force, given by τ  p  E , where p is the dipole moment of the dipole.
+6.00 C
–6.00 C
5 cm
5 cm
5 cm
14. Four point charges are arranged in the xy-plane as
sketched at right, two of magnitude +6.00 C and
two of magnitude –6.00 C, all a distance 5.0 cm
from the origin.
(a) Find the magnitude and direction of the
electric field at the origin.
5 cm
–6.00 C
Part III: Calculation: [60 points]
Choose three of the following four questions and
perform the indicated calculations (20 points each)
+6.00 C
The magnitude of the electric field from any one charge is just given by
Ei 
9
2
2
6
ke q  8.988 10 N  m /C  6.00  10 C 

 2.157 107 N/C
2
2
r
 0.05 m 
However, electric field is a vector quantity, and therefore we need to add the
contributions from all four charges. The electric field always points away from positive
and towards negative charges. It follows that the two charges to the right and left will
both produce a field pointing to the right, or the î direction, while the up and down
charges both produce a field in the up or ĵ direction. The total field will therefore be


 
E  2.157ˆi  2.157ˆi  2.157ˆj  2.157ˆj 107 N/C  4.314 ˆi  ˆj 107 N/C
The total magnitude of the electric field is then
E  Ex2  E y2 
 4.314 10
7
N/C    4.314  107 N/C   6.101107 N/C .
2
2
The direction is clearly in the first quadrant, at an angle given by
tan   Ex E y  1 , or   45
(b)A point charge of magnitude – 1.00 C and mass 1.00 g is placed at the
origin. Find the force on the charge (magnitude and direction) and the
acceleration of the point charge (magnitude and direction).
The force on a charge is given simply by F = qE, which works out to
F   1.00 106 C  6.101107 N/C   61.0 N
Since the sign came out negative, this is the opposite direction, or at an angle of
  180  45  225 . We can then use F = ma to get the acceleration, which will be in
the same direction (   225 ) and will have a magnitude of
a
F
61.0 N

 6.10 104 m/s 2
3
m 1.00 10 kg
conducter
3 cm
10.0 m
2 cm
2 cm
15. A long thin cylinder has a uniform
charge on it. It is 1.0 cm in radius and 10.0
m in length. It is then surrounded by a
neutral hollow conducting cylinder with
inner radius 2.0 cm and outer radius 3.0 cm,
also of length 10.0 m. It is observed that at a
distance of 4.0 cm from the axis of the
cylinder, the electric field is 50.0 kV/m,
oriented outwards. This field is nearly
constant along the entire length of the
cylinder.
(a) What is the total charge of the cylinder of charge?
cylinder
of charge
This, as worded, is a straight Gauss’s law application. We draw a cylinder of
radius 4.0 cm and length 10.0 m. We are told that the electric field at this radius is 50.0
kV/m. Because the electric field is outward, the total flux will be simply given by
 E  AE  nˆ  AE  2 RLE  2  0.0400 m 10.0 m   5.00 104 V/m 
 1.261105 V  m .
By Gauss’s Law, the total charge contained inside it must be
Q   0  E   8.854  1012 C2 /N/m 2 1.261105 V  m   1.112  106 C2  V/N/m
 1.112  C .
This is the total charge inside this hypothetical cylinder. Since the conductor is neutral
(it says so), and it is wholly contained inside this cylinder, it contributes nothing, and this
must all be attributed to the charged core.
(b) What is the electric field E a distance r = 2.5 cm from the axis (inside the
conductor)?
That’s easy; the electric field inside the conductor is always zero, also written 0.
(c) What is the total charge on the interior surface of the hollow conducting
cylinder?
If we draw another cylinder, this time with a radius of 2.5 cm, it is clear that since
there is no electric field, there is also no total charge inside this cylinder. The only places
there are charges are on the core (with charge 1.112  C ), and the inner surface of the
conductor, the conductor must have a cancelling charge of 1.112  C .
16. A ring of total charge Q = 2.00 C and radius a = 3
cm is arranged in the xy-plane, centered on the
origin.
(a) Find a formula for the potential V at a point on
the z-axis a distance z from the origin.
The distance between an arbitrary point on the ring and a
point on the z-axis can be found from the Pythagorean
theorem, which tells us
z
a = 3.0 cm
r 2  a 2  z 2 , so r  a 2  z 2
For each little piece, we now add up its charge divided by this distance, but the distance
is a constant, so the formula simply becomes
V 
ke dq ke
ke q
  dq 
.
r
r
a2  z2
We can then substitute in specific numbers as needed. In particular, for the next part of
the problem we will need it at z = 4.0 cm and z = 0.0 cm. For these two points, we have
V  z  0 cm  
kq
8.988 10

9
N  m 2 /C2  2.00 106 C 
0.030 m
a2
 5.99 105 N  m/C  599 kV ,
V  z  4 cm  
kq
8.988 10

9
N  m 2 /C2  2.00 106 C 
a2  z2
0.0302 +0.0402 m
 3.60 105 N  m/C  360 kV ,
(b) A particle of mass m = 3.00 g and charge q = 1.00 C is launched from z = 4.0
cm towards the center of the ring along the z-axis. Find the minimum
velocity v such that it can reach and pass through the center of the ring.
To reach the origin, the potential energy must increase by an amount
U  qV   5.99 105 V  3.60 105 V 1.00 106 C   0.239 J
The only source of such energy must be the initial kinetic energy of the charge. The
minimum speed then is found by using
1
2
mv 2  U ,
v2 
2  0.239 J 
2U

 478 m 2 /s 2 ,
3
m
1.00 10 kg
v  478 m 2 /s 2  21.9 m/s .
Technically, you need more than this speed to make it through the ring.
17. Three capacitors, initially uncharged, are connected to a switch and a battery as
shown at right. Two capacitors C1 and C2
C1
are of known capacitance, and a third C3
10.0 F
is of unknown capacitance. Initially, the
switch is open and the capacitors are
15.0 V
uncharged.
C3
(a) When the switch is closed, a total of
C2
750 C of charges flow out of the
15.0 F
battery and into the system of
capacitors. What must be the
effective capacitance of the whole combination of the three capacitors?
The combination of three capacitors can be summarized as a single capacitor. If
we call this effective capacitance C, then the capacitance can be found using
Q  C V ,
Q 750  C
C

 50.0  F .
V 15.0 V
(b) What is the capacitance of the unknown capacitor C3?
The capacitors C1 and C2 are in series, since they have one end joined and nothing
else joined to either of them. Therefore they can be treated as a single capacitor with
capacitance
1
1
1
1
1
3 2
5
1
 





,
C12 C1 C2 10.0  F 15.0  F 30.0  F 30.0  F 6.0  F
C12  6.0  F .
This is then in parallel with the capacitor C3. We therefore have
C  C3  C12 ,
C3  C  C12  50.0  F  6.0  F  44.0  F .
(c) What is the total amount of energy stored in the capacitors?
This can be worked out in a variety of ways. One easy one is
U  12 C  V  
2
1
2
 50.0 10
6
C  15.0 V   0.00563 J  5.63 mJ .
2