In the circuit shown in figure, magnitude of instantaneous power consumed by the capacitor is 

If Q be the instantaneous charge, then $\frac{\mathrm{Q}}{\mathrm{C}}=\mathrm{v}$

$\therefore \mathrm{Q}=\mathrm{Cv}={\mathrm{CV}}_0\mathrm{cos\omega t}\Rightarrow \mathrm{i}=\frac{\mathrm{dQ}}{\mathrm{dt}}=-{\mathrm{CV}}_0\mathrm{\omega sin\omega t}$

$\therefore$ Magnitude of instantaneous power

$\mathrm{p}=\left|\mathrm{vi}\right|={\mathrm{V}}_0\mathrm{cos\omega t}\times {\mathrm{CV}}_0\mathrm{\omega sin\omega t}=\frac{{\mathrm{V}}_0^2\mathrm{\omega C}}{2}\sin 2\mathrm{\omega t}$