A capacitor of capacitance C , is connected across an ac source of...

$∵ q = CV$

And $V = {V}_{0}sin\omegat$

${I}_{d} = ?$

As we know,

$∵ {I}_{d} = {ε}_{0}\dfrac{dQ}{dt}$

$= {ε}_{0}\dfrac{dEA}{dt}\left(∵ E = \dfrac{V}{d}\right)$

$= \dfrac{{ε}_{0}AdV}{ddt}$

$= \left(\dfrac{{ε}_{0}A}{d}\right)\dfrac{d}{dt}{V}_{0}sin\omegat$

${I}_{d} = C{V}_{0}\omega cos\omegat$