Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 44


36 V

Work Step by Step

$Energy=\frac{1}{2}CV^{2}$ Using the above equation, for capacitor B, we can write $3.4\times10^{-4}\,J=\frac{1}{2}C\times(12\,V)^{2}$ $\implies C=\frac{3.4\times10^{-4}\,J\times2}{(12\,V)^{2}}=4.7\times10^{-6}\,F$ As the capacitors are identical, C is same for A. Using the equation $E=\frac{1}{2}CV^{2}$, for capacitor A, we have $3.1\times10^{-3}\,J=\frac{1}{2}\times4.7\times10^{-6}\,F\times (V_{A})^{2}$ $\implies V_{A}=\sqrt {\frac{3.1\times10^{-3}\,J\times2}{4.7\times10^{-6}\,F}}=36\,V$
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