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by Eq. 214. When these two expressions are used in the equation
mg
tan
θ
=
F
e
, we
obtain
1/3
22
2
00
1
.
24
2
mgx
q
q L
x
Lx
m
g
εε
⎛⎞
≈⇒
≈
⎜⎟
ππ
⎝⎠
(b) We solve
x
3
= 2
kq
2
L
/
mg
for the charge (using Eq. 215):
()
3
2
3
8
92
2
0.010kg
9.8m s
0.050m
2.4 10 C.
2
2 8.99 10 N m C
1.20m
mgx
q
kL
−
==
=
±
×
×⋅
Thus, the magnitude is
8
2
.
41
0C
.
q
−
=×
54. (a) A force diagram for one of the balls is shown on the right.
The force of gravity
mg
G
acts downward, the electrical force
G
F
e
of
the other ball acts to the left, and the tension in the thread acts along
the thread, at the angle
to the vertical. The ball is in equilibrium,
so its acceleration is zero. The
y
component of Newton’s second
law yields
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 Spring '08
 Any
 Physics, Force, Gravity

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