Question

The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor x × 10-3 m. The value of x is,
(Permittivity of free space, E 0 = 8.85 × 10-12 C2 N-1 m-2)

Answer 1

The correct answer is 8
$4.425μ A = \frac{E_0A}{d} × \frac{dV}{dt}$
⇒ $d =\frac{8.85 × 10^{-12}× 40 × 10^{-4}}{4.425 × 10^{-6}} × 10^6$
⇒ d = 8 × 10–3 m
⇒ x = 8