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(A) When the battery is removed,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab o

Vedclass · Accepted Answer

The potential difference between the plates and the energy stored will decrease,but the charge on the plates will remain the same.

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(A) When the battery is removed,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab of dielectric constant $K > 1$ is inserted between the plates,the capacitance increases according to the formula $C' = KC$.Since $Q$ is constant and $C' = KC$,the new potential difference $V' = Q/C' = Q/(KC) = V/K$. Thus,the potential difference decreases.The energy stored in the capacitor is given by $U = Q^2 / (2C)$. Since $C$ increases,the energy stored $U' = Q^2 / (2C') = U/K$ decreases.Therefore,the potential difference and the energy stored decrease,while the charge remains the same.

After charging a capacitor,the battery is removed. Now,by placing a dielectric slab between the plates :-

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