Physics 121 Honors
Common Exam 1
Spring 2003
Name:
Section:
Closed book exam. Only one 8
.
5
00
×
11
00
formula sheet (front and back side) can be used. Calculators are
allowed. Use the scantron forms (pencil only!) for the multiple choice problems. Circle the answers on the
examination sheet as well, and return it together with the scantron form. Use the back of these pages, or
attach your own pages with solutions for problems that require calculations.
The multiplechoice problems are 5 points each. Partial credit (up to 3 points) will be given for multiple
choice problems, if you provide a detailed solution on the examination sheets. The workout problems are
10 points each. The nominal number of points is 100.
Clearly print your first and last name and indicate your section number on both the scantron form and the
examination sheet.
Good luck!
MultipleChoice Problems
Problem 1:
Two tiny conducting balls of identical mass
m
and identical charge
q
=
3
.
20
×
10

8
C hang
from nonconducting threads of length
L
=
140 cm. The angle in the equilibrium position as indicated in the
figure is
θ
=
2
.
8
◦
. What is the mass
m
of the conducting balls?
A)
m
=
1 g
B)
m
=
2 g
C)
m
=
10 g
D)
m
=
20 g
E)
m
=
40 g
F
e
=
T
sin
θ
mg
=
T
cos
θ
⇒
tan
θ
=
F
e
mg
and
sin
θ
=
x
2
L
⇒
m
=
q
2
4
πε
0
x
2
g
tan
θ
=
q
2
16
πε
0
gL
2
tan
θ
sin
2
θ
=
0
.
001 kg
Problem 2:
The radius of the nucleus of a Gold atom is about
r
Au
≈
1
.
4
×
10

15
3
√
A
m, where
A
=
197 u
is the atomic mass of Gold. What is the magnitude of the repulsive electrostatic force between two protons
separated by
r
Au
?
A)
F
=
0
.
60 N
B)
F
=
3
.
5 N
C)
F
=
8
.
4 N
D)
F
=
120 N
E)
F
=
1
.
8
×
10
5
N
F
=
1
4
πε
0
e
2
r
2
Au
=
e
2
4
πε
0
(
1
.
4
×
10

15
m
)
2
A
2
/
3
=
3
.
47 N
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Problem 3:
Three point charge lie along the
x
axis. The positive charge
q
1
=
12
.
0
μ
C is at
x
1
=
1
.
50 m,
the positive charge
q
2
=
5
.
20
μ
C is at the origin, and the resultant force on the negative charge
q
3
is zero.
What is the
x
coordinate of
q
3
?
1
4
πε
0
q
2
q
3
x
2
=
1
4
πε
0
q
1
q
3
(
x
1

x
)
2
⇒
√
q
2
(
x
1

x
)
=
√
q
1
x
x
=
√
q
2
√
q
1
+
√
q
2
x
1
= +
0
.
595 m
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 Spring '08
 Opyrchal
 Physics, Electric charge, Millikan, Rau

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