The charge on a parallel plate capacitor is varying as q = q0 sin 2πnt 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
$I_D= \frac{dq}{dt}=\frac{d}{dt} q_0 \sin 2 \pi nt=2 \pi nq_0 \cos 2 \pi q_0 \cos 2 \pi nt$