P22_015.PDF
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Chapter 22 - 22.15
15. (a) A force diagram for one of the balls is shown below. The force of gravity
mg
acts downward,
the electrical force
F
e
of the other ball acts to the left, and the tension in the thread acts along
the thread, at the angle
θ
to the vertical. The ball is in equilibrium, so its acceleration is zero.
The
y
component of Newton’s second law yields
T
cos
θ
−
mg
=0andthe
x
component yields
F
e
= 0. We solve the first equation for
T
and obtain
T
=
mg/
cos
θ
. We substitute the
result into the second to obtain
mg
tan
θ
−
−
F
e
=0.
y
.
.
T
θ
F
e
.
.
.
•
x
.
mg
Examination of the geometry of Figure 22-19 leads to
tan
θ
=
L
2
x/
2
.
−
(
x/
2)
2
x/
2
L
. This is equivalent to approximating tan
θ
by sin
θ
.The
magnitude of the electrical force of one ball on the other is
≈
F
e
=
q
2
4
πε
0
x
2
by Eq. 22-4. When these two expressions are used in the equation
mg
tan
θ
=
F
e
,weobtain
mgx
2
L
≈
1
4
πε
0
q
2
x
2
q
2
L
2
πε
0
mg
1
/
3
=
⇒
x
≈
.
(b) We solve
x
3
=2
kq
2
L/mg
) for the charge (using Eq. 22-5):
q
=
mgx
3
2
kL
=
(0
.
010kg)(9
.
8m
/
s
2
)(0
.
050m)
3
2(8
.
99
×
10
9
N
·
m
2
/
C
2
)(1
.
20m)
=
±
2
.
4
×
10
−
8
C
.
T
sin
θ
If
L
is much larger than
x
(which is the case if
θ
is very small), we may neglect
x/
2inthe
denominator and write tan
θ
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