The displacement current of $4.425 \mu \mathrm{~A}$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $10^{6} \mathrm{Vs}^{-1}$. The area of each plate of the capacitor is 40 $\mathrm{cm}^{2}$. The distance between each plate of the capacitor is $\mathrm{x} \times 10^{-3} \mathrm{~m}$. The value of x is ______. (Permittivity of free space, $\mathrm{E}_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Capacitance

The displacement current of

4.425 μ A

is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of

1 0^{6} Vs^{- 1}

. The area of each plate of the capacitor is 40

cm^{2}

. The distance between each plate of the capacitor is

x \times 1 0^{- 3} m

. The value of x is ______.
(Permittivity of free space,

E_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

)

Answer: 8

The displacement current in a capacitor is given by:

I_{d} = ϵ_{0} \frac{A}{d} \frac{d V}{d t}

where

A

is the area of the plates,

d

is the separation, and

\frac{d V}{d t}

is the rate of change of voltage.

Rearranging for

d

d = \frac{ϵ _{0} A}{I _{d}} \frac{d V}{d t}

Substituting the given values:

d = \frac{8.85 \times 1 0 ^{- 12} \times 4 \times 1 0 ^{- 3} \times 1 0 ^{6}}{4.425 \times 1 0 ^{- 6}} = 8 \times 1 0^{- 3} m = 8 mm

Therefore,

x = 8

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer