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: 50

Answer

$6.7\times10^{-7}C $.

Work Step by Step

Use the capacitance equation, Q = CV. The initial charge on the capacitor is $Q_i = C_iV$. After the mica is inserted, the capacitance increases by a factor of K, and the voltage is unchanged because the battery is still attached to the capacitor. The final charge on the capacitor can be found by $Q_f=C_fV$. $$\Delta Q = Q_f-Q_i= C_fV - C_iV $$ $$\Delta Q = KC_iV - C_iV=(K-1) C_iV=(7-1)(3500\times10^{-12}F)(32V)=6.7\times10^{-7}C $$
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