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: 14

Answer

The charge on the ball bearing is -10 nC.

Work Step by Step

Let $q_1$ be the charge on the glass bead. Let $q_2$ be the charge on the ball bearing. We can find the magnitude of $q_2$. $F = \frac{k~q_1~q_2}{r^2}$ $q_2 = \frac{F~r^2}{k~q_1}$ $q_2 = \frac{(0.018~N)(0.010~m)^2}{(9.0\times 10^9~N~m^2/C^2)(20\times 10^{-9}~C)}$ $q_2 = 10~nC$ The magnitude of the charge on the ball bearing is 10 nC. Since the ball bearing feels a downward force, the charge on the ball bearing must be negative. The charge on the ball bearing is -10 nC.
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