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 - General Problems - Page 499: 85

Answer

a) $6.4\times10^{-11}C$ b) $6.4\times10^{-11}C$ c) $18V$ d) $1.9\times10^{-10}J$

Work Step by Step

a) $Q=CV=(8.85\times10^{12}C^2/Nm^2)(\frac{3.0\times10^{-4}m^2}{0.5\times10^{-3}m})(12V)=6.4\times10^{-11}C$ b) After the capacitor is charged, moving the plates will not have an impact on its charge. $6.4\times10^{-11}C$ c) $C=(8.85\times10^{12}C^2/Nm^2)(\frac{3.0\times10^{-4}m^2}{0.75\times10^{-3}m})$$=3.54\times10^{-12}$ $V=\frac{Q}{C}=\frac{6.4\times10^{-11}C}{3.54\times10^{-12}}=18V$ d) $W=\frac{QV_{ba}}{2}=\frac{(6.4\times10^{-11}C)(18V-12V)}{2}=1.9\times10^{-10}J$
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