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To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 $\mu F$, the rate of change of applied variable potential difference $\left(\dfrac{dV}{dt}\right)$ must be:
A
800 V/s
B
500 V/s
C
200 V/s
D
400 V/s
Detailed Solution
$Q=CV\Rightarrow\dfrac{dQ}{dt}=C\dfrac{dV}{dt}\Rightarrow\dfrac{dV}{dt}=\dfrac{I}{C}=\dfrac{2\times10^{-3}}{4\times10^{-6}}=\dfrac{10^3}{2}=500$ V/s.