A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

Options

A. Reduction of charge on the plates and increase of potential difference across the plates
B. Increase in the potential difference across the plate, reduction in stored energy, but no charge in the charge on the plates
C. Decrease in the potential difference across the plates, reduction in the stored energy, but no charge in the charge on the plates
D. None of the above

Answer

C

Explanation

Battery in disconnected so `Q` will be constant as `C prop K`. So with introduction of dielectric slab capacitance will increase using `Q = CV, V` will decrease and using `U = (Q^(2))/(2C)`, energy will decrease.