Solution:
Since displacement current is given by \(I_d = C \frac{dV}{dt}\), we find \(\frac{dV}{dt} = \frac{I_d}{C} = \frac{2 \times 10^{-3} \text{A}}{4 \times 10^{-6} \text{F}} = 500 \text{V/s}\).
To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance \(4 \mu\text{F}\), the rate of change of applied variable potential difference \(\left(\frac{dV}{dt}\right)\) must be
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