Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - Problems - Page 497: 40

Answer

$3.39\times10^{-8}C$.

Work Step by Step

Combine equations 17–7 and 17–8 and find the charge. $Q=CV$ and $C=\frac{\epsilon_oA}{d}$ $$ Q=\frac{\epsilon_oA}{d}V$$ The desired electric field is V/d for the capacitor. $$=\epsilon_o(A)(E)$$ $$=(8.85\times10^{-12}C^2/(N\cdot m^2))(45.0\times10^{-4}m^2)(8.50\times10^5V/m)$$ $$=3.39\times10^{-8}C$$
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