Three concentric spherical shells have radii $a$, $b$ and $c\left(a<b<c\right)$ and have surface charge densities $+\sigma ,-\sigma ,+\sigma$ respectively. If $V_A,V_B$ and $V_C$ denote the potentials of the three shelIs, then for $c=a+b$, we have

It is clear from figure that

$V_A=\frac{\sigma a}{{\epsilon }_0}-\frac{\sigma b}{{\epsilon }_0}+\frac{\sigma c}{{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}(a-b+c)$

As $c=a+b$

$\begin{array}{r}\therefore V_A=\frac{\sigma }{{\epsilon }_0}\left(2a\right)\\ V_B=\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2}{b}-b+c\right]=\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2}{b}-b+a+b\right]\\ V_B=\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2}{b}+a\right]=\frac{\sigma a}{{\epsilon }_0}\frac{(a+b)}{b}=\frac{{\sigma }_{0}ac}{{\epsilon }_{0}b}\end{array}$

$\begin{array}{r}{V}_C=\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2}{c}-\frac{b^2}{c}+c\right]=\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2-b^2+c^2}{c}\right]\\ =\frac{\sigma }{{\epsilon }_0}\left[\frac{(a+b)(a-b)+(a+b{)}^2}{c}\right]\\ V_C=\frac{\sigma }{{\epsilon }_0}\left[\frac{(a+b)}{c}(a-b+a+b)\right]=\frac{\sigma 2a}{{\epsilon }_0}\end{array}$

We found that,$V_C=V_B=V_A$