Question

The displacement current of $4.425\mu A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $10^{6}V{s}^{-1}$. The area of each plate of the capacitor is $40c{m}^2$. The distance between each plate of the capacitor is $x\times 10^{-3}m$. The value of $x$ is, (Permittivity of free space, $E_0=8.85\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$ )

Answer 1

Answer: 8

Displacement Current $=$ Conduction Current
$=\frac{dq}{dt}$
$I_d=\frac{{ϵ}_{0}A}{d}\frac{dV}{dt}$
$d=\frac{8.85\times 10^{-12}\times 4\times 10^{-3}\times 10^6}{4.425\times 10^{-6}}$
$=8mm$
$X=8$