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 . 8 5 * 1 0 − 1 2 C 2 N − 1 m − 2 ) _ _ _ _ _ _ _ _ _ _

According to definition of displacement current, we can write I d = ε 0 d ? d t = ε 0 d ( E S ) d t = ε 0 d ( V l S ) d t = ε 0 S l ( d V d t ) ⇒ l = 8 . 8 5 * 1 0 − 1 2 * 4 0 * 1 0 − 4 * 1 0 6 4 . 4 2 5 * 1 0 − 6 8 * 1 0 − 3 m

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, $E_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$**

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Vishal Baghel | Contributor-Level 10

4 months ago

According to definition of displacement current, we can write

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

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

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