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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

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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).

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Solution

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,

`I_d = ε_0(dphi_E)/(dt)`

`I_d = ε_0 d/dt(EA)`

`I_d = (ε_0A)/d xx (dV)/(dt)`

`4.425 xx 10^-6 = (8.85 xx 10^-12 xx 40 xx 10^-4 xx 10^6)/d`

`d = (8.85 xx 40 xx 10^-10)/(4.425 xx 10^-6)`

`d = 8 xx 10^-3` m

Hence, x = 8

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