A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400V. If the plate area is 60 cm², then the value of displacement current for 10^–6 sec. will be:

1.062 × 10^-4 A

$I_{D}=\varepsilon_0 \frac{\partial \phi_{E}}{\partial t}=\varepsilon_0 \frac{EA}{t}=\frac{\varepsilon_0 VA}{dt}$

or  $I_{D}=\frac{8.85 \times 10^{-12} \times 400 \times 60 \times 10^{-4}}{2 \times 10^{-3} \times 10^{-6}}$

= 1.062 × 10^-4 A