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 30 - Electromagnetic Induction - Exercises and Problems - Page 871: 31


$C = 8.3\times 10^{-18}~F$

Work Step by Step

We can find the required capacitance: $f = \frac{1}{2\pi~\sqrt{LC}}$ $f^2 = \frac{1}{4\pi^2~LC}$ $C = \frac{1}{4\pi^2~L~f^2}$ $C = \frac{1}{(4\pi^2)~(15\times 10^{-3}~H)~(450\times 10^6~Hz)^2}$ $C = 8.3\times 10^{-18}~F$
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