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 498: 55

Answer

a) $C=9.08\times10^{-12}F$ b) $Q=8.17\times10^{-11}C$ c) $E=225\frac{V}{m}$ d) $E_P=3.68\times10^{-10}J$ 3) The answer to a, b, and d will change.

Work Step by Step

a) $C=\epsilon_o\frac{A}{d}=(8.85\times10^{-12}\frac{C^2}{N\cdot m^2})\frac{0.0410m^2}{0.04m}$ $=9.08\times10^{-12}F$ b) $Q=CV=(9.08\times10^{-11}F)(9V)=8.17\times10^{-11}C$ c) $E=\frac{V}{d}=\frac{9V}{0.04m}=225\frac{V}{m}$ d) $E_P=\frac{QV}{2}=\frac{(8.17\times10^{-11}C)(9V)}{2}=3.68\times10^{-10}J$ 3) If a dielectric is inserted, you increase the dielectric constant, increasing the capacitance. Changing the capacitance will change the charge and changing the charge will change the electric potential. The answer to a, b, and d will change.
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