JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 24)
A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is $7 \times 10^8 \mathrm{~V} / \mathrm{s}$ then the integer value of the distance between the parallel plates is
$\left(\right.$ Take, $\left.\epsilon_0=9 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}, \pi=\frac{22}{7}\right)$ ____________ $\mu \mathrm{m}$.Explanation
Given, $$r = 10\,cm = {1 \over {10}}\,m$$
$$I = 0.15\,A$$, and $${{dv} \over {dt}} = 7 \times {10^8}v/s$$
We know, for a parallel plate capacitor,
$$c = {{{\varepsilon _0}A} \over d} \Rightarrow c = {{{\varepsilon _0}\pi {r^2}} \over d}$$ .... (1)
In a capacitor,
$$Q = CV$$
$$ \Rightarrow V = {Q \over C}$$
by differentiating w.r.t. t
$$ \Rightarrow {{dv} \over {dt}} = {1 \over C}{{dQ} \over {dt}}$$
$$ \Rightarrow {{dv} \over {dt}} = {d \over {{\varepsilon _0}\pi {r^2}}}I$$ (As $$I = {{dQ} \over {dt}}$$)
$$ \Rightarrow d = {{{\varepsilon _0}\pi {r^2}} \over I}{{dv} \over {dt}}$$ .... (From (1))
$$ \Rightarrow d = {{9 \times {{10}^{ - 12}}} \over {0.15}} \times {{22} \over 7} \times {1 \over {100}} \times 7 \times 10$$
$$ = {{198} \over {0.15}} \times {10^{ - 6}}$$
$$ = 1320 \times {10^{ - 6}}$$ m
$$ \Rightarrow d = 1320\,\mu m$$
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