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 - Page 745: 12


(a) $F = 0.90~N$ (b) $a = 0.90~m/s^2$

Work Step by Step

(a) We can find the magnitude of the electric force on each mass: $F = \frac{1}{4\pi ~\epsilon_0}\frac{q_1q_2}{r^2}$ $F = \frac{1}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)}\frac{(10\times 10^{-6}~C)(10\times 10^{-6}~C)}{(1.0~m)^2}$ $F = 0.90~N$ (b) We can find the acceleration. $a = \frac{F}{m}$ $a = \frac{0.90~N}{1.0~kg}$ $a = 0.90~m/s^2$
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