Physics 101 Exam #3: Electricity
April 29, 2010
Three metal spheres A, B & C are shown below each charged as indicated.
A
+10 C
B
C
-4 C
-8 C
1. What will be the charge on each sphere after completing the following steps:
 Sphere A is brought into contact with Sphere C and then separated, then
 Sphere B is brought into contact with Sphere A and then separated, then
 Sphere A is brought into contact with Sphere C and then separated.
Step
Motion of Charges
Charge
Distribution
Initial Condition
None
A: +10 nC
B: -4 nC
C: -8 nC
Sphere A is brought
into contact with
Sphere C and then
separated,
-8 nC travel from Sphere C to Sphere A to neutralize all
but +2 nC on Sphere A. An additional -1 nC of charge
leaves Sphere C to neutralize all but +1 nC on Sphere A.
No more charges move since the charge on each sphere
is the same at +1 nC
A: +1 nC
B: -4 nC
C: +1 nC
Sphere B is brought
into contact with
Sphere A and then
separated
-1 nC travels from sphere B to sphere A to neutralize all
of sphere A. An additional -1.5 nC moves from Sphere
B to sphere A leaving -1.5 nC on both A and B.
A: -1.5 nC
B: -1.5 nC
C: +1 nC
Sphere A is brought
into contact with
Sphere C and then
separated.
-1 nC from sphere A moves to sphere C to neutralize it.
An additional -.25 nC moves from A to C so that each
sphere has a -.25 nC charge.
A: -.25 nC
B: -1.5 nC
C: -.25 nC
The final distribution of charge will be A: -.25 nC, B: -1.5 nC, and C: -.25 nC. To check this
result test to see if the net charge is the same before and after: initial net charge = +10-4-8=-2
nC, final net charge = -.25-1.5-.25 = -2 nC. Check
Will Sphere A, after completing the steps above, have excess electrons or a deficiency in
electrons? How many electrons will Sphere A have in excess or in deficiency?
Sphere A has a charge of -.25 nC so it has an excess of electrons. This corresponds to
.25  10 9 C
 1.56  10 9 electrons .
C
1.6  10 -19 e
You are provided an initially neutral
electroscope and a rod charged with
negative charges.
2. List the steps required to leave a net
positive charge on the electroscope.
Explain what charges are being affected
and how they are moving with each
step.
Step
-- -
-
Action
1
Bring negative rod near but not touching the top of the electroscope. This will polarize the
electroscope, pushing electrons away (toward the bottom) and leaving positive charges at the
top.
2
Touch the top of the electroscope with your finger allowing the electrons that were at the bottom
of the electroscope to escape to ground (even further away).
3
Remove your finger from the top of the electroscope leaving the electroscope deficient in
electrons and positively charges.
4
Withdraw the negative rod and the remaining electrons on the electroscope, through deficient
will spread out over the electroscope and leave a net negative charge.
Physicists believe that in the beginning of the Universe that there were precisely equal
numbers of protons and electrons created so that the Universe is neutral overall. The
following problem asks you to do a calculation that tests the notion that large
astronomical bodies are electrically neutral.
3. It can be shown that if there was even 1 excess electron for every trillion (1012) protons
that the Sun would have 1.01 x 1044 excess electrons and the Earth would have 1.79 x
1039 excess electrons. Given that the distance between these two bodies is 1.5 x 1011 m,
calculate the repulsive electric force between the Sun and Earth if there was even 1
excess electron per trillion (1012) protons.
This is a simple Coulomb’s Law problem. First the excess electrons on te Sun and Moon
must be converted into Coulombs of charge
C
 1.62  10 25 C
eC
1.79  10 39 electrons  1.79  10 39 e - 1.6  10 19 -  2.86  10 20 C
e
The force between the “charged” Sun and Earth will be
1.01 10 44 electrons  1.01 10 44 e - 1.6  10 19
2
25
q1q2 
C 2.86  10 20 C
9 N  m  1.62  10



9

10
 1.85 10 33 N
2
2


11
r2
C
1.5 10 m


The repulsive electric force between the Earth and Sun would be 1.85 x 1033 N.
F k
It can be further shown that the Earth is held in orbit around the Sun by the attractive
force of gravity with a magnitude of 3.53 x 1022 Newtons. Given your answer to the
hypothetical repulsive electric force between the Sun and Earth above, what would
happen to the Earth in its orbit if there were even 1 excess electron for every trillion
protons?
Since the repulsive electric force is many times larger than the attractive gravity force, I
would expect the Earth to fly off its orbit and be repelled into deep space.
A horizontally-oriented electric dipole
consisting of ±6 nC charges separated
by 6 cm is shown.
6 cm
+6 nC
4. What is the magnitude and direction of
the electric field 4 cm directly below the
dipole?
-6 nC
θ
r
4 cm
r
E-
The total electric field will be the vector
ENet
θ
sum of the electric fields from the
positive E+ and the negative E- charges.
E+
As you can see from the diagram, the
vertical components of these individual
electric fields will cancel since the
distances and charges are the same in magnitude. The net field will be directed to the
Q
right and will be given by E Net  2k 2  cos( ) . We need only determine r and θ.
r
r  a 2  b2 
tan( ) 
3 cm2  4 cm2
 5 cm
4 cm
 1.67    tan 1 1.67  59.1
3 cm
Calculate the net electric field
ENet  2k
2
9

Q
C
N
9 N  m  6 10



cos(

)

2
9

10
 cos(59.1 )  22,185
2
2
2


r
C  0.05 m
C

The net electric field at the point is 22,185 N/C directed towards the right.
5. What is the electric potential at the point 4 cm directly below the electric dipole?
The form for the electric potential is V (r )  k
Q
for a single point charge. For two point
r
charges we must add the individual potentials
Q
Q
1
1
VTotal (r )  k   k   k  Q  Q   k  0  0 . Thus the electric potential is zero
r
r
r
r
since the distance is the same from each charge and the charges are opposites of each
other.
A small sphere of mass 25 g hangs from a horizontal surface in the presence of an
electric field whose magnitude is 50,000 N/C directed to the left as shown. The sphere is
hanging at an angle of 30 degrees to the right.
6. What is the magnitude and polarity of the
charge on the sphere?
After constructing the free body diagram,
the force balance equations are
30
T
30
 
Vertically: T sin 30  FE
m
q
 
mg
Horizontally: T cos 30  mg
Combining
FE
 tan 30 
mg
 
FE
E

 
or FE  mg  tan 30 


Since the electric force is given as FE  qE , we can find q by simple division
 
 
mg  tan 30

E

.025 kg    9.8 m2   tan 30  
s 
 2.83  10 6 C
N
50,000
C
-6
The charge has magnitude of 2.83 x 10 C and must be negative in polarity. (Since the
electric force is in the opposite direction of the electric field.)
mg  tan 30   qE or q 

7. A 12 volt car battery can move 15 C of charge in 1 second to turn over an automobile’s
engine. How much work does the battery do if the car starts in 1 second?
This is an application of Work = qV. The Work done in one second by the battery will be
just 15 C x 12 Volts = 180 Joules. Recall a Volt is a J/C.
8. A distant proton has an initial velocity of 5 x 107 m/s is directed toward a 5 cm radius
sphere that carries a charge of 7.9 x 10-8 C. How close to the sphere will the proton
approach before being turned back by the electric repulsion?
+Q
vi
e+
This is an energy conservation problem where kinetic energy
electric potential energy PE  qV  q  k
1 2
mv in converted into
2
Q
.
r
Simple set the initial KE to the final PE and solve for r
KE  PE
1 2
Q
mv  q  k
2
r

N  m2 
  7.9  10 8 C
2  1.6  10 19 C   9  10 9
2
C 
2qkQ

r

 5.45  10 5 m
2
2
mv
m

1.67  10 27 kg   5  10 7 
s







The proton will approach to within 5.45 x 10-5 meters of the central charge.
Note: I made a typo in this problem. The particle approaches to closely to the spherical
charge. In fact it enters the spherical charge and the problem changes from a simple
point charge to something we did not study.
A nanoscale physicist has assembled 3 microscopic particles on a substrate. The
particles lie on the corners of a square that has sides of length 6 μm (1 μm = 10-6 m) as
shown in the figure. Each particle has a charge of 0.15 nC.
6 μm
q
q
6 μm
6 μm
q
6 μm
Substrate
9. What is the electric potential at the empty corner of the square?
The electric potential at the empty corner is just the sum of the individual electric
Q
potentials V (r )  k .
r
For the three charges shown the total electric potential at the empty corner is
VTotal (r )  V1 r1   V2 r2   V3 r3 
k
0.15 nC
0.15 nC
0.15 nC
k
k
6 m 
6 m  8.48 m 

N  m2 
1
1
1


  0.15  10 9 C  
  9  10 9


2
6
6
6
C 
 6  10 m 6  10 m 8.48  10 m 




N  m2 
1

  0.15  10 9 C  4.51 10 5 
  9  10 9
2
m
C 


 6.09  10 5 V


How much work is required to move another identical particle into the empty corner of
the square?
The work required to bring in the last charge will be just W = qV, or
W = (0.15 x 10-9 C) (6.09 x 105 V) = 9.14 x 10-5 Joules