Physics - EMI/AC Question with Solution | TestHub

From the given information,

$C=\frac{{\in }_0\mathrm{A}}{d}=\frac{{ϵ}_0\times 1}{2\times 10^{-3}}\mathrm{F}$
$\therefore X_C=\frac{1}{\omega \mathrm{C}}=\frac{2\times 10^{-3}}{2\times 50\pi \times {\in }_0}=\frac{2\times 10^{-3}}{25\times 4\pi {\in }_0}\Omega$
$\therefore X_C=\frac{2\times 10^{-3}}{25}\times 9\times 10^9=\frac{18}{25}\times 10^6\Omega$
$\therefore i_0=\frac{V_0}{X_C}=\frac{20\times 25}{18}\times 10^{-6}\mathrm{A}=27.47\mu \mathrm{A}.$

The value of amplitude of displacement current will be same as value of amplitude of conventional current.
Hence option $3.$