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A parallel plate capacitor without any dielectric within its plates, has a capacitance C, and is connected to a battery of emf V. The battery is disconnected and the plates of the capacitor are pulled apart until the separation between the plates is doubled. What is the work done by the agent pulling the plates apart, in this process ? |
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Answer» `1/2CV^(2)` where A is thearea of eachplateand dis the distancebetweenthe PLATES. Initialenergystoredin the capacitor , `U_(i) = (1)/(2) CV^(2) "".....(ii)` Whenthe separationbetweenthe plates is doubled , itscapcaitancebecomes `C'=(epsi_(0)A)/(2d)=(1)/(2) (epsi_(0)A)/(d) = (C)/(2) ""("USING (i)")"".....(iii)` As the bettaryis disconnected , socharged capacitorbecomes isolatedand charge on itwillreamain constant , `therefore"" Q = Q` `C'V = C'V ""("As" Q = CV)` `V= ((C)/(C))V =(C)/(((C)/(2)))V = 2V "".....(iv)` Finalenergystoredin the capacitor , `U_(f) = (1)/(2) C'V^(2) = (1)/(2) ((C)/(V)) (2V)^(2) = CV^(2)""...(V)` Requiredwork done , `W = U_(f) - U_(i) = CV^(2) - (1)/(2) CV^(2) = (1)/(2)CV^(2) = (1)/(2)CV^(2)` |
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