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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