A parallel plate capacitor is charged to a po | vedclass

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(A) When the capacitor is disconnected from the source of $emf$,the charge $Q$ on the plates remains constant because there is no path for the charge

Vedclass · Accepted Answer

only $(i)$ and $(ii)$

Vedclass · Answer

(A) When the capacitor is disconnected from the source of $emf$,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.Inserting a dielectric slab increases the capacitance $C$ of the capacitor,as $C = K C_0$,where $K > 1$ is the dielectric constant.Since the charge $Q$ is constant and the capacitance $C$ increases,the potential difference $V$ across the plates must change according to the relation $V = Q / C$. As $C$ increases,$V$ decreases.Therefore,the quantities that change are the potential difference $(i)$ and the capacitance $(ii)$. The charge $(iii)$ remains constant.

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