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

Easy

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

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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