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QJEEM22Phbc98: JEE Main 2022 Physics - Class 12 JEEM22Phbc98 Solution: The displacement current of 4.425 A is developed in the space…

The displacement current of 4.425 μ\mu 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 ×\times 10 −- 3 m. The value of x is __________.

(Permittivity of free space, E0 = 8.85 ×\times 10 −- 12 C2 N −- 1 m −- 2).

Step-by-Step Verified Solution

4.425 μA=E0Ad×dVdt4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}

⇒d=8.85×10−12×40×10−44.425×10−6×106\Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}

⇒d=8×10−3\Rightarrow d = 8 \times {10^{ - 3}} m

⇒x=8\Rightarrow x = 8

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