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

Published by Pearson

# Chapter 25 - Electric Charges and Forces - Exercises and Problems: 66

#### Answer

$q = +1.8\times 10^{-7}~C$

#### Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find $T_y$: $T_y = mg$ $T_y = (0.0050~kg)(9.80~m/s^2)$ $T_y = 0.049~N$ We can find the horizontal component of the tension $T_x$: $\frac{T_x}{T_y} = tan(\theta)$ $T_x = T_y~tan(\theta)$ $T_x = (0.049~N)~tan(20^{\circ})$ $T_x = 0.018~N$ The horizontal component of tension is equal in magnitude to the electric force on the ball. We can find the magnitude of the charge on the ball. $E~q = T_x$ $q = \frac{T_x}{E}$ $q = \frac{0.018~N}{100,000~N/C}$ $q = 1.8\times 10^{-7}~C$ Since the electric force on the ball is directed to the right, the charge on the ball is positive. $q = +1.8\times 10^{-7}~C$

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