The charge on a parallel plate capacitor varies as $\mathrm{q}=\mathrm{q}_0 \cos 2 \pi \mathrm{vt}$. The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor is:
Answer & explanation
Correct answer: option 2
Displacement current, $\mathrm{I}_{\mathrm{D}}=$ conduction current, $\mathrm{I}_{\mathrm{C}}$
∴ $\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{q}_0 \cos 2 \pi \mathrm{vt}\right]=-\mathrm{q}_0 2 \pi \mathrm{v} \sin 2 \pi \mathrm{vt}$