Q86 / 90 · JEE Main 2022 (29 Jun Shift 2) · Physics

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 V s-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=8.85×10-12 C2 N-1 m-2) _______

Correct answer: 8

Solution

Displacement current is given by id=ε0dϕEdt

Or id=ε0ddtEA, where E=qAε0

Or id=ε0ddtqAAε0=dqdt=ddtCV

Or id=CdVdt=ε0AddVdt

Putting the values, we have 

4.425×10-6=8.85×10-12×40×10-4×106 d

d=2×10-6×10-4×106×40

d=80×10-4=8×10-3 m

Hence, value of x=8.

Hence, the correct answer is 8.

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