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 m$$^2$$. The amplitude of the oscillating displacement current for the applied AC voltage is [Take $$\varepsilon_0 = 8.85 \times 10^{-12}$$ F m$$^{-1}$$]

The displacement current amplitude equals $$I_d = C \cdot \frac{dV}{dt}\bigg|_{max} = C \cdot V_0\omega$$, where $$C = \frac{\varepsilon_0 A}{d}$$ is the capacitance of the parallel plate capacitor.

The capacitance is: $$C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 1}{2 \times 10^{-3}} = 4.425 \times 10^{-9}$$ F.

The angular frequency is $$\omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314.16$$ rad s$$^{-1}$$.

The amplitude of the displacement current is: $$I_d = C V_0 \omega = 4.425 \times 10^{-9} \times 20 \times 314.16 = 4.425 \times 10^{-9} \times 6283.2 \approx 27.8 \times 10^{-6}$$ A $$= 27.79\,\mu$$A.