A charge $Q$ is distributed over three concentric spherical shells of radii $a,b,c\left(a<b<c\right)$ such that their surface charge densities are equal to one another. The total potential at a point at distance $r$ from their common centre, where $r<a$, would be :

The correct option is B

$\frac{Q\left(a+b+c\right)}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{a}+{\mathrm{b}}^2+{\mathrm{c}}^2\right)}$

Step 1 : Given data

Charge $=Q$

Radii of three spherical shell $=a,b,c$

The Potential at the point $P$

Here the surface charge densities are equal to one another

Step 2 : To find the total potential at the point

, $$\begin{gathered} V=\frac{k{Q}_A}{a}+\frac{k{Q}_b}{b}+\frac{k{Q}_c}{c} \\ {Q}_a:Q_b:Q_c=a^2:b^2:c^2(\sin ce{\sigma }_{a:}{\sigma }_b:{\sigma }_c) \\ {Q}_a=\left[\frac{a^2}{a^2+b^2+c^2}\right]Q \\ {Q}_b=\left[\frac{b^2}{a^2+b^2+c^2}\right]Q \\ {Q}_c=\left[\frac{c^2}{a^2+b^2+c^2}\right]Q \\ V=\frac{Q}{4\pi {\epsilon }_0}\left[\frac{\left(a+b+c\right)}{a^2+b^2+c^2}\right] \end{gathered}$$

Hence the total potential $V=\frac{Q\left(a+b+c\right)}{4{\mathrm{\pi \epsilon }}_0\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)}$

Therefore the correct answer is Option B