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 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. The value of x is,

(Permittivity of free space, $E_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$

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Vishal Baghel | Contributor-Level 10

2 weeks ago

According to definition of displacement current, we can write

$I_{d} = ε_{0} \frac{d ?}{d t} = ε_{0} \frac{d (E S)}{d t} = ε_{0} \frac{d (\frac{V}{l} S)}{d t} = \frac{ε_{0} S}{l} (\frac{d V}{d t})$

$\Rightarrow l = \frac{8.85 \times 1 0^{- 12} \times 40 \times 1 0^{- 4} \times 1 0^{6}}{4.425 \times 1 0^{- 6}} 8 \times 1 0^{- 3} m$

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