For a parallel plate capacitor of capacitance 2µF to establish an instantaneous displacement current of 2.0 mA in the space between its plates, the rate of change of voltage required is:

10³ V s^-1

The correct answer is Option (4) → 10³ V s^-1

The formula for displacement current ($I_d$) is -

$I_d=C\frac{dV}{dt}$

$⇒\frac{dV}{dt}=\frac{I_d}{C}=\frac{2×10^{-3}}{2×10^{-6}}=10^3Vs^{-1}$