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(A) The charge on the capacitor is given by $q = CV$.Substituting the given voltage $V = V_{0} \sin \omega t$,we get $q = C V_{0} \sin \omega t$.The d

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$I_{d} = V_{0} \omega C \cos \omega t$

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(A) The charge on the capacitor is given by $q = CV$.Substituting the given voltage $V = V_{0} \sin \omega t$,we get $q = C V_{0} \sin \omega t$.The displacement current $I_{d}$ is equal to the rate of change of charge on the capacitor plates,which is $I_{d} = \frac{dq}{dt}$.$I_{d} = \frac{d}{dt} (C V_{0} \sin \omega t)$.Since $C$ and $V_{0}$ are constants,$I_{d} = C V_{0} \frac{d}{dt} (\sin \omega t)$.Using the derivative of $\sin \omega t$,which is $\omega \cos \omega t$,we get $I_{d} = C V_{0} \omega \cos \omega t$.Therefore,$I_{d} = V_{0} \omega C \cos \omega t$.

$A$ capacitor of capacitance $C$ is connected across an $AC$ source of voltage $V$,given by $V = V_{0} \sin \omega t$. The displacement current between the plates of the capacitor would then be given by:

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