College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 18 - Problems - Page 698: 7


$I = 2.2\times 10^{-2}~A$

Work Step by Step

Since the positive charges and negative charges move in opposite directions, we can add the magnitudes of the flow of charges to find the total current. $I = (1.6\times 10^{-19}~C)[(2)(3.8\times 10^{16})+6.2\times 10^{16}]~s^{-1}$ $I = 2.2\times 10^{-2}~A$
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