In an AC circuit containing only capacitance the current

In A C circuit with capacitor

$E=E_o\sin \omega t\to 1$

$I=I_o\sin \left(\omega t+\frac{\pi }{2}\right)\to 2$

From Eq(1) and Eq(2) 

Current leads the voltage by  $\frac{{\pi }^C}{2}\left(or\right){90}^0$