Question

The displacement current of 4.425 µA is developed in the space between the plates of the 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 is x × 10^-3 m. The value of x is ______.

(Permittivity of free space, ε₀ = 8.85 × 10^-12C²N^-1m^-2).

विकल्प

• 6

• 7

• 8

• 9

Answer 1

The displacement current of 4.425 µA is developed in the space between the plates of the 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 is x × 10^-3 m. The value of x is 8.

Explanation:

As the displacement current,

`I_d = ε_0(dphi_E)/(dt)`

`I_d = ε_0 d/dt(EA)`

`I_d = (ε_0A)/d xx (dV)/(dt)`

`4.425 xx 10^-6 = (8.85 xx 10^-12 xx 40 xx 10^-4 xx 10^6)/d`

`d = (8.85 xx 40 xx 10^-10)/(4.425 xx 10^-6)`

`d = 8 xx 10^-3` m

Hence, x = 8