A displacement current of 1.0 A is established in the space between the two parallel plates of 1 μF capacitor. The rate of change of voltage across the capacitor is:

$10^6 V s^{-1}$

The correct answer is Option (1) → $10^6 V s^{-1}$

The displacement current ($I_D$) is related to the rate of charge of voltage $(dv/dt)$ by -

$I_D=C\frac{dv}{dt}$ [C → capacitance]

given,

$I_D=1.0A$

$C=1μF=1×10^{-6}F$

$∴\frac{dv}{dt}=\frac{I_D}{C}=\frac{1.0A}{1×10^{-6}}$

$\frac{dv}{dt}=10^6V/s$