Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 20 - Electric Circuits - Problems - Page 572: 5


(a) $1.5\times10^{-11}\,A$ (b) $4.7\times10^{7}\,ions$

Work Step by Step

(a) According to Ohm's law, $I=\frac{V}{R}=\frac{75\times10^{-3}\,V}{5.0\times10^{9}\,\Omega}=1.5\times10^{-11}\,A$ (b) $I=\frac{q}{t}\implies q=It=1.5\times10^{-11}\,A\times0.50\,s$ $=7.5\times10^{-12}\,C$ Let the number of $Na^{+}$ ions be n. Then, $n=\frac{q}{+e}=\frac{7.5\times10^{-12}\,C}{1.6\times10^{-19}\,C/ion}=4.7\times10^{7}\,ions$
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