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: 49

Answer

9.5 volts

Work Step by Step

The capacitor is disconnected from the battery, so the charge on it stays the same. However, the capacitance is increased by a factor of the dielectric constant, $\kappa$. Let $Q_i$ and $V_i$ denote the initial charge and voltage on the capacitor, and let $Q_f$ and $V_f$ denote the final charge and voltage. Equation 17–7, Q=CV, relates the charges and voltages to the capacitance. $$Q=C_iV_i=C_fV_f$$ $$V_f=V_i\frac{C_i}{C_f}= V_i\frac{C_i}{\kappa C_i}$$ $$V_f= (21.0V)\frac{1}{2.2}=9.5V$$
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