Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Exercises and Problems - Page 738: 21


(a) $C = 1.25\times 10^{-11}~F$ (b) $Q = 1.25\times 10^{-9}~C$

Work Step by Step

(a) We can find the capacitance: $C = \frac{\epsilon_0~A}{d}$ $C = \frac{\epsilon_0~\pi~r^2}{d}$ $C = \frac{(8.854\times 10^{-12}~F/m)(\pi)~(0.015~m)^2}{0.50\times 10^{-3}~m}$ $C = 1.25\times 10^{-11}~F$ (b) We can find the magnitude of the charge $Q$ on each electrode: $Q = C~\Delta V$ $Q = (1.25\times 10^{-11}~F)(100~V)$ $Q = 1.25\times 10^{-9}~C$
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