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Energy per unit volume for a capacitor having area A and separation d kept at potential diffeence V is given by : -A. `1/2 epsi_(0)(V^(2))/(d^(2))`B. `(1)/(2epsi_(0)) (V^(2))/(d^(2))`C. `1/2CV^(2)`D. `(Q^(2))/(2C)`

Answer» Correct Answer - A
Energy density `=1/2 epsi_(0)E^(2)`
Since `E=V/d`
Therefore energy density `=1/2epsi_(0)((V_(2))/(d_(2)))`


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