College Physics (4th Edition)

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

Chapter 18 - Problems - Page 698: 6

Answer

$I = 8.0\times 10^{-3}~A$

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. $I = (1.6\times 10^{-19}~C)(3.8\times 10^{16}+1.2\times 10^{16})~s^{-1}$ $I = 8.0\times 10^{-3}~A$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.