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:

(1) $I_d = V_0 \omega C cos \omega t $

(2) $I_d = \frac {V_0}{\omega C} cos \omega t $

(3) $I_d = \frac {V_0}{\omega C} sin \omega t$

(2) $I_d = \frac {V_0}{\omega C} cos \omega t $

(3) $I_d = \frac {V_0}{\omega C} sin \omega t$

(4) $I_d = V_0 \omega C sin \omega t $

*Solution*

For capacitor, $\frac {dq}{dt}$ is the displacement current and not the usual conduction current that exists in wire. This is because the plates of the capacitor are separated by some distance not allowing charge to flow in a manner as it does in a wire.

We have, $q=CV=CV_0 sin \omega t$

$I_d = \frac {dq}{dt} = CV_0 \omega cos \omega t$

*Alternate Solution*

In A.C. capacitive circuit, $I=\frac {V_0 sin (\omega t + \frac {\pi}{2})}{X_C}$

$\Rightarrow I=\omega C V_0 cos \omega t$

Answer: (1)