The displacement current of 4.425μ A is… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

2022integerhard

The displacement current of $4.425\mu A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of ${10}^{6}V{s}^{-1}$ . The area of each plate of the capacitor is $40{\mathrm{cm}}^{2}$ . The distance between each plate of the capacitor is $x\times {10}^{-3}m$ . The value of $x$ is ,

(Permittivity of free space, ${\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2}$ ) _______

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Held on 29 Jun 2022 · Verified 6 Jul 2026.

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Solution

Displacement current is given by ${i}_{d}=$ $\frac{{\epsilon }_{0}d{\phi }_{E}}{dt}$

Or ${i}_{d}=\frac{{\epsilon }_{0}d}{dt}(EA)$ , where $E=\frac{q}{A{\epsilon }_{0}}$

Or ${i}_{d}={\epsilon }_{0}\frac{d}{dt}(\frac{qA}{A{\epsilon }_{0}})=\frac{dq}{dt}=\frac{d}{dt}(CV)$

Or ${i}_{d}=C\frac{dV}{dt}=\frac{{\epsilon }_{0}A}{d}\frac{dV}{dt}$

Putting the values, we have

$4.425\times {10}^{-6}=\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}\times {10}^{6}}{d}$

$d=2\times {10}^{-6}\times {10}^{-4}\times {10}^{6}\times 40$

$d=80\times {10}^{-4}=8\times {10}^{-3}m$

Hence, value of $x=8$ .

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