Instantaneous displacement current of 1.0 A in the space between the parallel plates of 1µF capacitor can be established, by rate of change of potential difference:
Answer & explanation
Correct answer: option 2
$I_{d}=\varepsilon_0 \frac{d \phi_{E}}{dt}=\varepsilon_0 A \frac{d}{dt}\left(\frac{V}{d}\right)$
or $I_{d}=\frac{\varepsilon_0 A}{d} \times \frac{dV}{dt}=C \frac{dV}{dt}$
or $\frac{dV}{dt}=\frac{I_{d}}{C}=\frac{1.0}{10^{-6}}=10^6 Vs^{-1}$