Exam 1
PHY 2140
Row:
Seat:
Date, 2003
Please type your name here
Please type your student number here
Instructions:
The following procedure must be followed in order to correct any (unlikely) mistakes in
the grading.
1. Do all the questions. Show your work for partial credit. I must be able to
understand what you have done while I am reading the exam – not when you
explain it to me after the exam is graded and returned.
2. For each problem:
(a) write down any formula(s) used
(b) justify why you used that particular formula(s)
(c) copy the answer you have circled from the exam pages
to the General Purpose Answer Sheet and fill in the ovals.
(d) show your work!
3. The formula sheet is on the last page of the exam. You may detach it from the rest
of the exam.
NAME
1. Two identical conducting spheres are placed with their centers 0.30 m apart.
Initially, the first sphere has a charge of 12 x 10-9 C and the second a charge of
-18 x10-9 C. The spheres are then connected by a conducting wire. What is the
magnitude of the electrostatic force between the two spheres when the charges are
in equilibrium?
a.
b.
c.
d.
e.
3.0 x 10-7 N
5.0 x 10-7 N
6.0 x 10-7 N
9.0 x 10-7 N
not enough information
2. Three charges are arranged as shown, producing a net electric force on the charge
at the origin. Find the angle made by this force with the positive x axis.
yy
a.
b.
c.
d.
e.
37
45
130
225
260
--22 C
C
11 m
m
--88 C
C
xx
++22 C
C
22 m
m
.
3. Determine the point nearest the charges where the total electric field is zero. Take
the -2.50 C charge to be at the origin of the x axis.
11..0000 m
m
a.
b.
c.
d.
e.
–1.80 m
–1.27 m
0.27 m
0.39 m
1.27 m
--22..5500 C
C
--66..0000 C
C
NAME
4. The potential difference across two charged parallel plates is 200 V. This
potential difference causes a charged dust particle to move from one plate to
another. If the dust particle has charge 2.00 C and mass 1.00 x 10-5 kg, find the
speed of the dust particle as it reaches the second plate.
a.
b.
c.
d.
e.
0.73 m/s
8.94 m/s
160 m/s
3.2 x 104 m/s
Dust has no charge.
5. The plates of a parallel plate capacitor are separated by 1.00 x 10–4 m. If the
material in between them is a jelly with a dielectric constant of 2.26, what is the
plate area needed to provide a capacitance of 1.50 pF?
a.
b.
c.
d.
e.
1.00 x 10-6 m2
4.35 x 10-6 m2
7.50 x 10-6 m2
9.00 x 10-6 m2
I don’t think I’m ready for that jelly.
6. Find the charge on the 2 F capacitor in the following circuit.
a.
b.
c.
d.
e.
4 C
12 C
18 C
24 C
not enough information
22 FF
44 FF
22 FF
1122 V
V
7. A rectangular gold bar has two sides of length 0.04 m and one side of length 0.10
m. To prevent theft, the block can be connected to a 1.90 V source across any
opposite faces of the block. Find the current that can flow through the block along
its longer side. The resistivity of gold is 2.44 x 10-8 m.
a.
b.
c.
d.
e.
1.23 x 106 A
3.17 x 106 A
7.78 x 106 A
1.46 x 107 A
Resistance is futile.
8. The filament of a light bulb has a resistance of 20.0  at 20 C (when the light is
off and the filament is at room temperature). When the light is on, the resistance is
160 . Find the temperature of the filament when it is hot. Take the temperature
coefficient for resistivity to be 3.5 x 10-3 for the filament at 20 C.
a.
b.
c.
d.
e.
939 C
1120 C
1440 C
2020 C
2310 C
9. If the size of the charge value is tripled for both of two point charges maintained
at a constant separation, the mutual force between them will be changed by which
factor?
a.
b.
c.
d.
e.
9.0
3.0
1/3
1/9
Fear factor.
10. Consider a simple parallel-plate capacitor whose plates are given equal and
opposite charges and are separated by a distance d. Suppose the plates are pulled
apart until they are separated by a distance D>d. The electrostatic energy stored in
the capacitor is
a. greater than
b. the same as
c. smaller then
before the plates were pulled apart.
11. A “free” electron and a “free” proton are placed in an identical electric field. After
the particles are released, the acceleration of the proton will be
a. greater then
b. the same as
c. smaller than
the acceleration of the electron.
12. (bonus) Each of the protons in a particle beam has a kinetic energy of 3.25x10-15 J
What are the magnitude and direction of the electric field that will stop these
protons in a distance of 1.25 m?
a.
b.
c.
d.
e.
5.12 x 104 N/C and opposite to the motion
1.63 x 104 N/C and opposite to the motion
5.12 x 104 N/C and along the motion
1.63 x 104 N/C and along the motion
2.16 x 104 N/C and perpendicular to the motion
v
A
x f  xi
a
t f  ti
Ax  Ay
2
W  ( F cos  ) s
KE 
q1 q 2
E
2
energy 
P  I 2R 
1 2
mv
2
F
q0
q
r
1
1
1



Ceq C1 C 2
A
d
1
QV
2
R
Ax
V  ke
VB  VA  Ed
C   0
Ay
v v
x 0
t
 2 
2
r
t f  ti
tan  
2
v 2  v0  2ax
F  ke
v f  vi

A
V 2
R
energy 
Q2
2C
Ax  A cos 
x  v0 t 
Fx = m ax
Fy = m ay
Wnet = KEf - KEi
KEf +PEf = KEi + PEi
E  ke
PE  k e
q
r2
q1 q 2
r
C eq  C1  C 2  
I
Q
t
R  R0 1   T  T0 
Req  R1  R2  
1
1
1



Req R1 R2
ke=8.99 x 109 Nm2/C2
1 kWh = 3.60 x 106 J
1 2
at
2
v  v0  at
   0 1   T  T0 
g = 9.8 m/s2
 = 8.85 x 10-12 C2/Nm2
Ay  A sin 
1 eV = 1.6 x 10-19 J
PE  qVB  VA 
Q  CV
energy 
1
2
C V 
2
V  IR
P  IV