Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 627: 62


$\theta = 4.4^{\circ}$

Work Step by Step

The vertical component of the tension in each thread is equal in magnitude to the weight of one charge: $F_T~cos~\theta = mg$ $F_T = \frac{mg}{cos~\theta}$ Since $\theta$ is a small angle, then $sin~\theta \approx \theta$ and $tan~\theta \approx \theta$ We can find the angle $\theta$: $F_T~sin~\theta = \frac{kq^2}{r^2}$ $(\frac{mg}{cos~\theta})~(sin~\theta) = \frac{kq^2}{(2~L~sin~\theta)^2}$ $tan~\theta~sin^2~\theta = \frac{kq^2}{4L^2~mg}$ $\theta^3 = \frac{kq^2}{4L^2~mg}$ $\theta = (\frac{kq^2}{4L^2~mg})^{1/3}$ $\theta = [\frac{(9.0\times 10^9~N~m^2/C^2)(100\times 10^{-9}~C)^2}{(4)(1.0~m)^2~(0.0050~kg)(9.8~m/s^2)}]^{1/3}$ $\theta = 0.077~rad$ $\theta = (0.077~rad)(\frac{180^{\circ}}{\pi~rad})$ $\theta = 4.4^{\circ}$
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