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

Published by Pearson

# Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 624: 17

#### Answer

The net force on charge A is zero.

#### Work Step by Step

We can find the magnitude of the electric force on A from B: $F = \frac{k~q_A~q_B}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.010~m)^2}$ $F = 9.0\times 10^{-5}~N$ A force of $9.0\times 10^{-5}~N$ is directed to the right. We can find the magnitude of the electric force on A from C: $F = \frac{k~q_A~q_C}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(4.0\times 10^{-9}~C)}{(0.020~m)^2}$ $F = 9.0\times 10^{-5}~N$ A force of $9.0\times 10^{-5}~N$ is directed to the left. Since the same magnitude of force is directed to the right and left, the net force on charge A is zero.

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