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. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x×10−3m. The value of x is,(Permittivity of free space, E0=8.85×10−12C2N−1m−2)

Displacement Current = Conduction Current Id=dqdtId=ϵ0AddVdt⇒d=8.85×10−12×4×10−3×1064.425×10−6⇒d= 8×10−3 m⇒x=8

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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 $10^{6} \frac{V}{s}$ . The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is $x \times 10^{- 3} m$ . The value of x is,
(Permittivity of free space, $E_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )

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answer is 8.

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Detailed Solution

Displacement Current = Conduction Current
$\begin{array}{l} I_{d} = \frac{dq}{dt} \\ I_{d} = \frac{ϵ_{0} A}{d} \frac{dV}{dt} \\ \Rightarrow d = \frac{8 .85 \times 10^{- 12} \times 4 \times 10^{- 3} \times 10^{6}}{4 .425 \times 10^{- 6}} \\ \Rightarrow d = 8 \times 10^{- 3} m \\ \Rightarrow x = 8 \end{array}$