Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 23 - Gauss' Law - Problems - Page 681: 33a

Answer

$E=0$

Work Step by Step

According to Equation (23-13), the magnitude of the electric field due to the plate of positive charge is $E = \frac{\sigma}{2~\epsilon_0}$ and this electric field is directed away from the positively charged plate. According to Equation (23-13), the magnitude of the electric field due to the plate of negative charge is $E = \frac{\vert \sigma \vert}{2~\epsilon_0}$ and this electric field is directed toward the negatively charged plate. To the left of the plates, the electric field due to the negatively charged plate is directed to the right while the electric field due to the positively charged plate is directed to the left. Since the magnitude of the surface charge density is equal for both plates, the magnitude of each electric field due to each plate is equal. Therefore, by superposition, $E = 0$
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