The voltage between the plates of a parallel plate capacitor of capacitance 1.0 µF is changing at rate of 1.0 V/s. The displacement current is:

10 µA

Given C = 1.0 µF = 10⁻⁶ F, $\frac{dV}{dt}$ = 1 V/s

$\mathrm{I}_D=\varepsilon_0 \frac{\mathrm{d} \phi_E}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{VA}}{\mathrm{d}}\right)$

$=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}=1 \times 10^{-6}=10 \mu \mathrm{A}$