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 26 - The Electric Field - Exercises and Problems - Page 776: 17


The diameter of each disk is 18.6 mm

Work Step by Step

We can find magnitude of the charge that was transferred. $Q = (3.0\times 10^9)(1.6\times 10^{-19}~C)$ $Q = 4.8\times 10^{-10}~C$ We can find the area of each disk. $E = \frac{Q}{A~\epsilon_0}$ $A = \frac{Q}{E~\epsilon_0}$ $A = \frac{4.8\times 10^{-10}~C}{(2.0\times 10^5~N/C)(8.85\times 10^{-12}~C^2/N~m^2)}$ $A = 2.71\times 10^{-4}~m^2$ We can find the radius of each disk. $\pi~r^2 = A$ $r = \sqrt{\frac{A}{\pi}}$ $r = \sqrt{\frac{2.71\times 10^{-4}~m^2}{\pi}}$ $r = 0.0093~m = 9.3~mm$ Since the diameter is twice the radius, the diameter of each disk is 18.6 mm
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