A parallel plate capacitor is charged to a po | vedclass

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(A) When a 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 f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) When a 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.$(i)$ The capacitance $C$ of a parallel plate capacitor increases when a dielectric slab is inserted,given by $C = K C_0$,where $K > 1$ is the dielectric constant.$(ii)$ Since the charge $Q$ remains constant and the capacitance $C$ increases,the potential difference $V$ across the plates,given by $V = Q/C$,must decrease.$(iii)$ The charge $Q$ remains constant as the capacitor is isolated.Therefore,the potential difference $(i)$ and the capacitance $(ii)$ change,while the charge $(iii)$ remains constant. Thus,only $(i)$ and $(ii)$ change.

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