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  4. The displacement current of 4.425 μ A is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106Vs -1. The area of each plate of the capacitor is 40 cm 2. The distance between each plate of the capacitor is x × 10-3 m. The value of x is, (Permittivity of free space, E 0=8.85 × 10-12 C 2 N -1 m -2 )

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Q. 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 $10^{6}Vs ^{-1}$. The area of each plate of the capacitor is $40\, cm ^{2}$. The distance between each plate of the capacitor is $x \times 10^{-3} m$. The value of $x$ is, (Permittivity of free space, $E _{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}$ )

JEE MainJEE Main 2022Electrostatic Potential and Capacitance

Solution:

Displacement Current $=$ Conduction Current
$=\frac{ dq }{ dt }$
$I_{d}=\frac{\epsilon_{0} A }{ d } \frac{ dV }{ dt }$
$d=\frac{8.85 \times 10^{-12} \times 4 \times 10^{-3} \times 10^{6}}{4.425 \times 10^{-6}}$
$=8\, mm$
$X =8$