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:
Answer & explanation
Correct answer: option 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,
$I_D=1.0A$
$C=1μF=1×10^{-6}F$
$∴\frac{dv}{dt}=\frac{I_D}{C}=\frac{1.0A}{1×10^{-6}}$
$\frac{dv}{dt}=10^6V/s$