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(A) When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates,the charge $(Q)$

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Potential difference and energy stored

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(A) When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates,the charge $(Q)$ remains constant because the circuit is open.The capacitance $(C)$ increases as $C = KC_0$,where $K$ is the dielectric constant.The potential difference $(V)$ decreases as $V = \frac{V_0}{K}$ because $V = \frac{Q}{C}$ and $C$ increases while $Q$ is constant.The stored energy $(U)$ decreases as $U = \frac{Q^2}{2C} = \frac{U_0}{K}$ because $C$ increases while $Q$ is constant.Therefore,the potential difference and the stored energy decrease.

$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

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