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 22 - Electric Charges and Forces - Exercises and Problems - Page 624: 8


$9.0\times 10^{13}~electrons$

Work Step by Step

We can find the total charge in 2 hours: $Q = (-2.0\times 10^{-9}~C/s)(2~h)(3600~s/h)$ $Q = -1.44\times 10^{-5}~C$ We can find the number of electrons: $N = \frac{-1.44\times 10^{-5}~C}{-1.6\times 10^{-19}~C} = 9.0\times 10^{13}~electrons$
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