To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 $μ$ F, the rate of change of applied variable potential difference $(\frac{dV}{dt})$ must be:-

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400 V/s

200 V/s

800 V/s

500 V/s

answer is B.

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Detailed Solution

Displacement current,
$i_{D} = C \cdot \frac{dV}{dt}$
where C is the capacitance
so, $\frac{dV}{dt} = \frac{i_{D}}{C} = \frac{2 \times 10^{- 3}}{4 \times 10^{- 6}} = 500 V / s$

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