The charge on a parallel plate capacitor is varying as $q = q_0 sin 2\pi ft$ The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:
Answer & explanation
Correct answer: option 3
$I_d = \frac{dq}{dt} = \frac{d}{dt} q_0 sin 2\pi ft = q_0 2\pi f cos 2\pi ft$