JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 24)
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
8
Explanation
$$4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}$$
$$ \Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}$$
$$ \Rightarrow d = 8 \times {10^{ - 3}}$$ m
$$ \Rightarrow x = 8$$
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