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(C) When the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric s

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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

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(C) When the battery is disconnected,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,the capacitance $C$ increases according to the relation $C' = KC$.Since $Q = CV$,the potential difference $V$ across the plates is given by $V = Q/C$. As $C$ increases and $Q$ remains constant,the potential difference $V$ decreases.The energy stored in the capacitor is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the stored energy $U$ decreases.Therefore,the potential difference decreases,the stored energy decreases,and the charge remains unchanged.

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

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