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: 45

Answer

$Q_1=1.53\times10^{-3}C$ $Q_2=4.16\times10^{-3}C$ $V=611V$

Work Step by Step

Equations needed: $Q=CV$ (pg 483) $Q_1=C_1V_1=(2.50\mu F)(746V)=1870\mu C$ $Q_2=C_2V_2=(6.8\mu F)(562V)=3820\mu C$ $\sum Q=1870\mu C+3820\mu C=5690\mu C$ Once the two capacitors are connected, there will be no voltage difference, it will all reach equilibrium, so $V_1=V_2=V_f$ $Q_1+Q_2=5690\mu C$ $\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ $C_2Q_1=C_1Q_2$ $C_2 5690\mu C-C_2Q_2=C_1Q_2$ $(6.8\mu F)(5690\mu C)-(6.8\mu F)Q_2=(2.5\mu F)Q_2$ $Q_2=4.16\times10^{-3}C$ $Q_1=1.53\times10^{-3}C$ $V=\frac{1.53\times10^{-3}C}{2.5\mu F}=611V$
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