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 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).
Answer (integer)
8
Solution
<p>$4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}$</p>
<p>$$ \Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}$$</p>
<p>$\Rightarrow d = 8 \times {10^{ - 3}}$ m</p>
<p>$\Rightarrow x = 8$</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Maxwell's Equations
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