Physics 101 Quiz Solutions
1. Two rectangular parallel metal plates are separated by a small distance. How many
electrons must be transferred from one plate to the other to give each plate a charge of
±157 nC?
157 nC  157 10 -9 C 
157  10 -9 C
 9.811011 e
C
1.6 10 -19
e
981 billion electrons must be transferred from one plate to the next to place 157
nC on charge on each plate.
2. A large metal sphere is
given a negative
charge. Two smaller
metal spheres, labeled
A and B, are mounted
on insulating stands
and are free to move.
A flexible grounding
wire is also available.
See figure. Describe in
a series of steps how
you would
_ _ _
_ _ _ _
_ _ _
A
B
Ground
a. Give both
spheres a negative charge.
Simply touch each small sphere to the larger sphere and some of the excess
negative charges will transfer to each smaller sphere.
A
_ _ _
_
_ _ _
_ _
_ B
b. Give one sphere a positive charge and the other sphere a negative charge.
While the small spheres are far from
___
the large negative sphere touch them
____
together.
___
Bring the two smaller spheres (still
touching each other) close to the large
negative sphere. The nearer sphere
will be polarize positive and the
farther sphere will be polarized
negative.
Separate the small spheres from each
other while close to the large negative
sphere.
The polarization is now locked into
each sphere.
c. Give both spheres a positive charge
While the small spheres are far from
the large negative sphere touch them
together.
Bring the two smaller spheres (still
touching each other) close to the large
negative sphere. The nearer sphere
will be polarize positive and the
farther sphere will be polarized
negative.
Ground the farther small sphere,
allowing the negative charges to run to
ground leaving the two smaller
spheres with a positive charge.
___
____
___
++ _ _
___
____
___
++
± ±
__
___
____
___
++
___
____
___
__
± ±
___
____
___
++ _ _
___
____
___
+
+
_
_
Disconnect the ground wire and
separate the small spheres from each
other while close to the large negative
sphere.
The polarization is now locked into
each sphere.
___
____
___
+
+
___
____
___
+
+
3. Two charges of -5nC and + 15nC are separated by a distance of 0.25 cm. What is the
magnitude and direction of the electric force between them?
qq
This is a straight forward Coulomb’s Law problem F  k 1 2 2 . Since the charges are of
r
opposite polarity the force will be radialy attractive.



2
9
q1q2 
C  15 10 9 C
9 N  m  5  10


F  k 2   9 10
 0.1 N
r
C 2 
0.0025 m2

The force between the two charges is 0.1 Newtons.
A note is worthy here: Many times the electrical force between charges may not seem
like a large number. Frequently the forces between charges will be very small especially
when dealing with individual electrons, protons or ions. However, the masses of those
particles are also typically very small – on the order of 10-25 to 10-31 kg. Thus even very
small forces can cause large accelerations and high speeds in these circumstances.
4. A small sphere, charged to -7 nC, is suspended from a fine
thread hanging at an angle of 30 degrees to the right in a
uniform electric field of 50,000 N/C.
a. What is the direction of the electric field? Sketch it on
the diagram.
E must be directed to the left since the force on the
charge is directed towards and right AND the charge is


a negative charge. F  qE
30

E
b. Sketch the free body diagram of forces on the small sphere.
30
T 30
FElec
mg
c.
Derive and calculate the mass of the sphere?
The net force on the sphere is zero. Balance the forces in the horizontal and vertical
Horizontally:
T∙sin(30) = FElec
Vertically:
T∙cos(30) = mg
tan( 30  ) 
FElec qE

mg mg
Combining
qE
m 

g  tan( 30  )
7 10

N

C   50,000 
C

 6.19  10 5 kg
m
9.8 2  tan( 30  )
s
9
The mass of the sphere is 61.9 milligrams
5. A light weight 2.5 g Styrofoam sphere carries a charge of 375 nC. It is placed between
two horizontal parallel plates carrying equal and opposite charges sit on a lab bench and
create a uniform electric field between them. The electric field between the plates is set
so that the Styrofoam sphere will “float” between the plates.
a. What is the magnitude of the electric field required to “float” the Styrofoam
sphere?
The sphere will “float” when the downward weight (mg) is balanced by an
upward electric force (qE).
Therefore mg = qE, or
mg
E

q
2.5 10

m

kg   9.8 2 
N
s 

 65,333
9
C
375  10 C
3
The electric field strength is 65,3333 N/C and directed upward.
b. Which of the parallel plates has a positive charge? The upper or lower? Explain
your choice.
The lower plate must be positive while the upper plate is negative for the E field to be
irected upwards.
6. An electric dipole is oriented along the y-axis as shown in the figure. The upper portion
of the dipole is +Q and the lower portion is –Q. An electron is placed on the x-axis as
shown.
a. What is direction of the NET electric field from the dipole at the electron’s
position? Justify your answer with a sketch and explanation. No calculations are
needed.
+
+ ½a
-e
E+
E-
- ½a
_
ENET
The electric field at the electron’s position is show due to each of the dipole
charges (Positive charge radial away, negative charge radial toward). You can see
from the symmetry of the problem that the horizontal components of each electric
field must cancel and the vertical components must add. Thus the Net E field is
directed downward, as shown.
b. In what direction will the electron initially accelerate if released from its position?
The force on the electron from a downward directed E field will be upward. The
electron will initially accelerate upward.