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, $\varepsilon_0 = 8.85 \times 10^{–12} C^2 N^{–1} m^{–2}) \)

Answer 1

$4.425\mu A=\frac{{\epsilon }_{0}A}{d}\times \frac{dV}{dt}$

$\Rightarrow d=\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}}{4.425\times {10}^{-6}}\times {10}^6$

$\Rightarrow d=8\times {10}^{-3}m$

$\Rightarrow x=8$