Assertion : A parallel plate capacitor is con | vedclass

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(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant $(V = V_0)$.The energy s

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If the Assertion is correct but Reason is incorrect.

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(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant $(V = V_0)$.The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.When a dielectric slab of dielectric constant $K$ is introduced between the plates,the capacitance becomes $C' = KC$.Therefore,the new energy stored is $U' = \frac{1}{2} (KC) V^2 = K U$. Thus,the energy stored becomes $K$ times.Since $Q = CV$,the charge on the plates becomes $Q' = KCV = KQ$. The surface charge density is given by $\sigma = \frac{Q}{A}$. Since $Q$ increases by a factor of $K$,the surface charge density $\sigma' = \frac{KQ}{A} = K\sigma_0$ also increases by a factor of $K$.Therefore,the Reason is incorrect because the surface charge density does not remain constant.

Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.Reason : The surface density of charge on the plate remains constant or unchanged.