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In order to establish an instantaneous displacement current of 1 mA in the space between the plates of 2 $μ$ F parallel plate capacitor, the rate of change of potential difference is

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$100 {Vs}^{- 1}$

$200 {Vs}^{- 1}$

$300 {Vs}^{- 1}$

$500 {Vs}^{- 1}$

answer is D.

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Detailed Solution

$I_{D} = 1 mA = 10^{- 3} A; C = 2 μ F = 2 \times 10^{- 6} F$

Since, $I_{D} = C . \frac{dV}{dt}$ $\Rightarrow \frac{dV}{dt} = \frac{I_{D}}{C} = 500$
Therefore, applying a varying potential difference of 500 Vs^-1 would produce a displacement current of desired value

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