Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 25 - Electric Charges and Forces - Exercises and Problems: 67

Answer

$\theta = 14.3^{\circ}$

Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find $T_y$: $T_y = mg$ $T_y = (0.0020~kg)(9.80~m/s^2)$ $T_y = 0.0196~N$ The horizontal component of tension $T_x$ is equal in magnitude to the electric force on the ball. We can find $T_x$: $T_x = E~q$ $T_x = (200,000~N/C)(25\times 10^{-9}~C)$ $T_x = 0.0050~N$ We can find the angle $\theta$: $tan(\theta) = \frac{T_x}{T_y}$ $\theta = arctan(\frac{T_x}{T_y})$ $\theta = arctan(\frac{0.0050~N}{0.0196~N})$ $\theta = 14.3^{\circ}$
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