Anderson, Matthew – Homework 2 – Due: Feb 5 2008, noon – Inst: Vitaly
1
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printout
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have
12
questions.
Multiplechoice questions may continue on
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be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 20 points
Forces between two objects which are in
versely proportional to the square o± the dis
tance between the objects include which o±
the ±ollowing?
I) Gravitational ±orce between two celestial
bodies
II) Electrostatic ±orce between two electrons
III) Nuclear ±orce between two neutrons.
1.
I only
2.
I, II, and III
3.
I and II only
correct
4.
II and III only
5.
III only
Explanation:
I) Gravitational ±orce
∝
1
r
2
.
II) Electrostatic ±orce
∝
1
r
2
.
III) The nuclear ±orce is very strong com
pared to the gravitational ±orce and electro
static ±orce over a very short range; outside
o± that range, however, it is negligible and
cannot be proportional to
1
r
2
keywords:
002
(part 1 o± 1) 20 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the fgure. The length o± the strings are equal
and the angle (shown in the fgure) with the
vertical is identical.
The acceleration o± gravity is 9
.
8 m
/
s
2
and
the
value
o±
Coulomb’s
constant
is
8
.
98755
×
10
9
N m
2
/
C
2
.
0
.
08
m
3
◦
0
.
01 kg
0
.
01 kg
Find the magnitude o± the charge on each
sphere.
Correct answer: 6
.
3301
×
10

9
C.
Explanation:
Let :
L
= 0
.
08 m
,
m
= 0
.
01 kg
,
and
θ
= 3
◦
.
L
a
θ
m
m
q
q
From the right triangle in the fgure above,
we see that
sin
θ
=
a
L
.
There±ore
a
=
L
sin
θ
= (0
.
08 m) sin(3
◦
)
= 0
.
00418688 m
.
The separation o± the spheres is
r
= 2
a
=
0
.
00837375 m
.
The ±orces acting on one o± the
spheres are shown in the fgure below.
θ
θ
m
g
F
T
e
T
sin
θ
T
cos
θ
Because the sphere is in equilibrium, the
resultant o± the ±orces in the horizontal and
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2
vertical directions must separately add up to
zero:
X
F
x
=
T
sin
θ

F
e
= 0
X
F
y
=
T
cos
θ

mg
= 0
.
From the second equation in the system
above, we see that
T
=
mg
cos
θ
, so
T
can be
eliminated from the ±rst equation if we make
this substitution. This gives a value
F
e
=
mg
tan
θ
= (0
.
01 kg)
(
9
.
8 m
/
s
2
)
tan(3
◦
)
= 0
.
00513596 N
,
for the electric force.
From Coulomb’s law, the electric force be
tween the charges has magnitude

F
e

=
k
e

q

2
r
2
,
where

q

is the magnitude of the charge on
each sphere.
Note:
The term

q

2
arises here because the
charge is the same on both spheres.
This equation can be solved for

q

to give

q

=
s

F
e

r
2
k
e
=
s
(0
.
00513596 N) (0
.
00837375 m)
2
(8
.
98755
×
10
9
N m
2
/
C
2
)
=
6
.
3301
×
10

9
C
.
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 Spring '08
 Opyrchal
 Charge, Force, Work, Correct Answer, Electric charge, noon – Inst

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