The instantaneous voltage, $$V = V_{0}sin (omega t)$$                    ...(i) Let $$q$$ be the charge on capacitor and $$I$$, the current in the circuit at any instant, then instantaneous potential difference, $$V=dfrac{q}{c}$$                        ...(ii) From (i) and (ii) $$dfrac{q}{C} = V_{0}sin(omega t) implies q = CV_{0} sin(omega t)$$ The instantaneous current, $$I = dfrac {dq}{dt} = (CV_{0}sin(omega t))=CV_{0}dfrac{d}{dt}(sin(omega t))=CV_{0}omega cos(omega t)$$ $$I=dfrac{V_{0}}{1/omega C} cos(omega t)$$ $$I=I_{0}sin(omega t + dfrac{pi}{2})$$ Hence, the current leads the applied voltage in phase by $$dfrac{pi}{2}$$