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Derive an expression for the energy stored in a charged parallel plate capacitor. |
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Answer» SOLUTION :Let at a particular instant charge on the plate of CAPACITOR be q and its POTENTIAL difference be `q/C.` If an additional charge dq is GIVEN to the capacitor plate, work done for it is given by `dW=(q/C).dq` Therefore, whole process of charging from 0 to Q requires a work `W = int_0^Q (qdp)/C = 1/C [q^2/2]_0^Q = Q^2/(2C)` This work done is stored as the electrostatic potential ENERGY of the charged capacitor. Hence, potential energy of charged capacitor `u = Q^2/(2C)` But Q=CV, where V be the potential difference between the plates of capacitur, hence `u = (Q^2)/(2C) =1/2QV =1/2 CV^2` |
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