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: 13

Answer

(a) $F = 56.3~N$ (b) The force is 287 times the weight of the sphere.

Work Step by Step

(a) We can find the magnitude of the electric force on each sphere: $F = \frac{1}{4\pi ~\epsilon_0}\frac{q_1q_2}{r^2}$ $F = \frac{1}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)}\frac{(50.0\times 10^{-9}~C)(50.0\times 10^{-9}~C)}{(0.020~m)^2}$ $F = 56.3~N$ (b) We can find the ratio of the force to the weight of one sphere. $\frac{F}{mg} = \frac{56.3~N}{(0.020~kg)(9.80~m/s^2)} = 287$ The force is 287 times the weight of the sphere.
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