Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 63

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ We know that the stored energy in the parallel-plate capacitor is given by $$U=\dfrac{Q^2}{2C}$$ where $C=A\epsilon_0/d$ Hence, $$U=\dfrac{Q^2d}{2A\epsilon_0}\tag 1$$ Plug the known; $$U=\dfrac{(10\times 10^{-9})^2(1\times 10^{-3})}{2(0.1^2)(8.85\times 10^{-12})}$$ $$U=\color{red}{\bf 5.65\times 10^{-7}}\;\rm J$$ $$\color{blue}{\bf [b]}$$ Now the only variable changed in (1) after pulling the plates away from each other is $d$ (the separation distance). All other variables remain constant. Hence, the stored energy is now given by $$U'=\dfrac{Q^2d'}{2A\epsilon_0} $$ Plug the known $$U'=\dfrac{(10\times 10^{-9})^2(2\times 10^{-3})}{2(0.1^2)(8.85\times 10^{-12})}$$ $$U'=\color{red}{\bf 11.3\times 10^{-7}}\;\rm J$$ $$\color{blue}{\bf [c]}$$ The energy is conserved when the system is isolated but our system is not isolated since we did some work on the system by pulling the plates away from each other. This is the reason for the difference.
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