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: 16

Answer

$E = 1.1\times 10^5~N/C$

Work Step by Step

We can find the electric field strength 2.0 mm above the center of the electrode. $E = \frac{\sigma}{2~\epsilon_0}$ $E = \frac{Q}{2~\epsilon_0~A}$ $E = \frac{8.0\times 10^{-8}~C}{(2)(8.85\times 10^{-12}~C^2/N~m^2)(0.20~m)(0.20~m)}$ $E = 1.1\times 10^5~N/C$
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