In an a.c. circuit with pure capacitance $C$ and a.c. source $E=E_0 \sin \omega t$,the equation of instantaneous current is given by

• A
  $I=E_0 \omega C \sin (\omega t)$
• B
  $I=E_0 \omega C \sin \left(\omega t+\frac{\pi}{2}\right)$
• C
  $I=\frac{E_0}{\omega C} \sin (\omega t)$
• D
  $I=\frac{E_0}{\omega C} \sin \left(\omega t+\frac{\pi}{2}\right)$