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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 625: 40


(a) The force on object A is 0.45 N (b) $q_A = 1.0\times 10^{-6}$ $q_B = 5.0\times 10^{-7}~C$

Work Step by Step

(a) Since both objects experience the same magnitude of electric force, the force on object A is also 0.45 N (b) Let $q$ be the charge on object B. Then the charge on object A is $2q$. We can find the charge $q$. $F = \frac{k~q_A~q_B}{r^2}$ $F = \frac{k~(2q)~q}{r^2}$ $2q^2 = \frac{F~r^2}{k}$ $q^2 = \frac{F~r^2}{2k}$ $q = \sqrt{\frac{F}{2k}}~r$ $q = \sqrt{\frac{0.45~N}{(2)(9.0\times 10^9~N~m^2/C^2)}}~(0.10~m)$ $q = 5.0\times 10^{-7}~C$ Since $q_A = 2q$, then $q_A = 1.0\times 10^{-6}$ and $q_B = 5.0\times 10^{-7}~C$.
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