QJEEM25Phadc5: JEE Main 2025 Physics - Class 12 JEEM25Phadc5 Solution: A parallel plate capacitor consisting of two circular plates of …

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

$\Rightarrow d = 1320\,\mu m$