Question

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

• 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 change in the charge on the plates
• C. Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
• D. None of the above

Answer 1

Answer: (C)

Battery in disconnected so $Q$ will be constant. As $C \propto K$, so with introduction of dielectric slab capacitance will increase. Using $Q=C V, V$ will decrease and using $U=\frac{Q^{2}}{2 C}$, energy will decrease.