Q 12-08-160JEE MainJEE Main 2021 (20 Jul, Shift 1)Medium
AC voltage $V(t) = 20\sin\omega t$ of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is $1\ \text{m}^2$. The amplitude of the oscillating displacement current for the applied AC voltage is [Take $\varepsilon_0 = 8.85\times10^{-12}\ \text{F m}^{-1}$]
Answer: (C) 27.79 $\mu$A
$C = \dfrac{\varepsilon_0A}{d} = \dfrac{8.85\times10^{-12}}{2\times10^{-3}} = 4.425\times10^{-9}$ F.
The displacement current equals the charging current $C\dfrac{dV}{dt}$, with amplitude
$$I_0 = CV_0\omega = 4.425\times10^{-9}\times20\times2\pi\times50 \approx 2.78\times10^{-5}\ \text{A} = 27.79\ \mu\text{A}$$
Solution by Sreeraj P, M.Sc Physics