JEE Main 2019 — Electrostatics Question with Solution

Surface charge densities of all sphere is given as $\frac{q_1}{a^2}=\frac{q_2}{b^2}=\frac{q_3}{c^2}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$, where $q_1$, $q_2$ and $q_3$ is charge on sphere of radius $a$, $b$ and $c$ respectively.

$q_1+q_2+q_3=Q$

$q_1=\frac{a^2\left(Q\right)}{a^2+b^2+c^2}$

$q_2=\frac{b^2\left(Q\right)}{a^2+b^2+c^2}$

$q_3=\frac{c^2\left(Q\right)}{a^2+b^2+c^2}$

Potential at a point where $r<a$ is given as $V=\frac{K{q}_1}{a}+\frac{K{q}_2}{b}+\frac{K{q}_3}{c}$, where $K$ is a constant.

$V=\frac{1}{4\pi {\epsilon }_0}\left[Q\right]\left(\frac{a+b+c}{a^2+b^2+c^2}\right)$