Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 27 - Current and Resistance - Exercises and Problems - Page 762: 13


$I = 1.8\times 10^{-8}~A$

Work Step by Step

We can find the average current: $I = \frac{Q}{t}$ $I = \frac{9.0\times 10^{-12}~C}{0.50\times 10^{-3}~s}$ $I = 1.8\times 10^{-8}~A$
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