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: 23


$L = 4.8~cm$

Work Step by Step

We can find the value of $L$: $C = \frac{\epsilon_0~A}{d}$ $C = \frac{\epsilon_0~L^2}{d}$ $L^2 = \frac{C~d}{\epsilon_0}$ $L = \sqrt{\frac{C~d}{\epsilon_0}}$ $L = \sqrt{\frac{(100\times 10^{-12}~F)(0.20\times 10^{-3}~m)}{8.854\times 10^{-12}~F/m}}$ $L = 0.048~m$ $L = 4.8~cm$
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