756_Physics ProblemsTechnical Physics

756_Physics ProblemsTechnical Physics - 96 Capacitance and...

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96 Capacitance and Dielectrics P26.58 (a) We use Equation 26.11 to find the potential energy of the capacitor. As we will see, the potential difference V changes as the dielectric is withdrawn. The initial and final energies are U Q C i i = F H G I K J 1 2 2 and U Q C f f = F H G I K J 1 2 2 . But the initial capacitance (with the dielectric) is CC if = κ . Therefore, U Q C f i = F H G I K J 1 2 2 . Since the work done by the external force in removing the dielectric equals the change in potential energy, we have WU U Q C Q C Q C fi iii =− = F H G I K J F H G I K J = F H G I K J 1 2 1 2 1 2 1 222 κκ af . To express this relation in terms of potential difference V i , we substitute QC V ii =∆ bg , and evaluate: WC V = × = × −− 1 2 1 1 2 200 10 100 500 100 400 10 2 9 2 5 a f ej a f a f .. . . F V J . The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work done on the system by the external force that pulled out the dielectric.
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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