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)

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

Answer 2

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