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

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

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

Reduction of charge on the plates and increase of potential difference across the plates

Increase of charge on the plates and increase of potential difference across the plates

answer is C.

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Detailed Solution

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 $U = \frac{Q^{2}}{2 C}$ , energy will decrease.

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