## 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: 17

#### Answer

The net electric force on charge B is $4.05\times 10^{-4}~N$ directed downwards.

#### Work Step by Step

We can find the magnitude of the electric force on charge B from charge A: $F = \frac{k~q_A~q_B}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.020~m)^2}$ $F = 4.5\times 10^{-5}~N$ A force of $4.5\times 10^{-5}~N$ is directed down. We can find the magnitude of the electric force on charge B from charge C: $F = \frac{k~q_B~q_C}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(2.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.010~m)^2}$ $F = 3.6\times 10^{-4}~N$ A force of $3.6\times 10^{-4}~N$ is directed down. Since both forces are directed down, we can add the forces to find the net force. $F_{net} = 4.5\times 10^{-5}~N+3.6\times 10^{-4}~N$ $F_{net} = 4.05\times 10^{-4}~N$ The net electric force on charge B is $4.05\times 10^{-4}~N$ directed downward.

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