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Homework 4 Solutions
Spring 2010
Required Problems
Chapter 21
8.
Use the charge per electron and the mass per electron.
(
29
(
29
6
14
14
19
31
14
16
1 electron
46 10 C
2.871 10
2.9 10 electrons
1.602 10
C
9.109 10
kg
2.871 10 e
2.6 10
kg
1 e







×
=
×
≈
×

×
×
×
=
×
12.
Let the right be the positive direction on the line of charges.
Use the fact that like charges
repel and unlike charges attract to determine the direction of the forces.
In the following
expressions,
9
2
2
8.988 10 N m
C
k
=
×
⋅
.
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
75
2
2
48
2
2
85
2
2
75 C
48 C
75 C
85 C
ˆ
147.2 N
150 N
0.35m
0.70m
75 C
48 C
48 C
85 C
ˆ
ˆ
ˆ
ˆ
563.5N
560 N
0.35m
0.35m
85 C
75 C
85 C
48 C
ˆ
416.3N
420 N
0.35m
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
k
k
k
k
k
k
μ
+
+

= 
+
= 
≈ 
=
+
=
≈
= 

= 
≈ 
F
i
F
i
i
i
i
F
i
i
i
i
i
i
i
r
r
r
14.
(
a
)
If the force is repulsive, both charges must be positive since the total charge is
positive.
Call the
total charge
Q
.
(
29
(
29
(
29
(
29
(
29
(
29
2
2
1
1
1
2
1
2
1
1
2
2
2
2
2
2
1
2
2
6
6
1
2
9
2
2
0
4
4
2
2
12.0N
1.16m
90.0 10 C
90.0 10 C
4
8.988 10 N m C
kQ Q Q
kQ Q
Fd
Q
Q
Q
F
Q
QQ
d
d
k
Fd
Fd
Q
Q
Q
Q
k
k
Q



+
=
=
=
→

+
=
±

±

=
=
=
×
±
×

×
⋅
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6
60.1 10 C , 29.9 10 C


=
×
×
(
b
)
If the force is attractive, then the charges are of opposite sign.
The value used for
F
must then
be negative.
Other than that, the solution method is the same as for part (
a
).
(
29
(
29
(
29
(
29
(
29
(
29
2
2
1
1
1
2
1
2
1
1
2
2
2
2
2
2
1
2
2
6
6
1
2
9
2
2
6
6
0
4
4
2
2
12.0N
1.16m
90.0 10 C
90.0 10 C
4
8.988 10 N m C
106.8 10 C ,
16.8 10 C
kQ Q Q
kQ Q
Fd
Q
Q
Q
F
Q
QQ
d
d
k
Fd
Fd
Q
Q
Q
Q
k
k
Q





+
=
=
=
→

+
=
±

±

=
=

=
×
±
×

×
⋅
=
×

×
20.
If all of the angles to the vertical (in both cases) are assumed to
be small, then the spheres only have horizontal displacement,
and so the electric force of repulsion is always horizontal.
Likewise, the small angle condition leads to
tan
sin
θ
≈
≈
for
all small angles.
See the freebody diagram for each sphere,
showing the three forces of gravity, tension, and the
electrostatic force.
Take to the right to be the positive
horizontal direction, and up to be the positive vertical direction. Since the spheres are in
equilibrium, the net force in each direction is zero.
(
a
)
1
T1
1
E1
E1
T1
1
sin
0
sin
x
F
F
F
F
F
=

=
→
=
∑
1
1
1
T1
1
1
T1
E1
1
1
1
1
1
1
1
cos
sin
tan
cos
cos
y
m g
m g
F
F
m g
F
F
m g
m g
=

→
=
→
=
=
=
∑
A completely parallel analysis would give
E2
2
2
F
m g
=
.
Since the electric forces are a
Newton’s third law pair, they can be set equal to each other in magnitude.
E1
E2
1
1
2
2
1
2
2
1
1
F
F
m g
m g
m m
θ θ
=
→
=
→
=
=
(
b
)
The same analysis can be done for this case.
E1
E2
1
1
2
2
1
2
1
1
2
F
F
m g
m g
m m
=
→
=
→
=
=
(
c
)
The horizontal distance from one sphere to the other is
s by the small angle approximation.
See the diagram.
Use the
relationship derived above that
E
F
mg
=
to solve for the distance.
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 Spring '08
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