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).

पर्याय

• 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,

${\displaystyle I_d={\epsilon }_0\frac{d{\varphi }_E}{dt}}$

${\displaystyle I_d={\epsilon }_0\frac{d}{dt}\left(EA\right)}$

${\displaystyle I_d=\frac{{\epsilon }_{0}A}{d}\times \frac{dV}{dt}}$

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

${\displaystyle d=\frac{8.85\times 40\times {10}^{-10}}{4.425\times {10}^{-6}}}$

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

Hence, x = 8