The charge of a parallel plate capacitor is varying as $q= q_0 sin \, 2\pi ft$ . The plates are very large and close together (area= A, separation= d). Neglecting edge effects, the displacement current through the capacitor is
Answer & explanation
Correct answer: option 3
$i = \frac{dq}{dt} =\frac{d}{dt} (q_0 sin 2 \pi ft) = q_02\pi f cos 2 \pi ft $