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JEE Main 2022 · 29 Jun · Shift 2, Physics question 56

The displacement current of 4.425 μ\muA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10610^6 Vs−1^{-1}. The area of each plate of the capacitor is 40 cm2^2. The distance between each plate of the capacitor is x×10−3x\times10^{-3}m. The value of xx is ______

(Permittivity of free space, E0=8.85×10−12E_0=8.85\times10^{-12} C2^2 N−1^{-1} m−2^{-2})

Answer and worked solution

Official answer: 8

Displacement current I=ε0AddVdtI=\dfrac{\varepsilon_0A}d\dfrac{dV}{dt}: d=8.85×10−12×40×10−4×1064.425×10−6=8×10−3d=\dfrac{8.85\times10^{-12}\times40\times10^{-4}\times10^6}{4.425\times10^{-6}}=8\times10^{-3} m.

Answer: 8

Topics:Displacement current
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