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

The correct option is B $V_C$ = $V_A\ne V_B$
It is clear from figure that
$V_a=\frac{\sigma a}{{\epsilon }_0}-\frac{\sigma b}{{\epsilon }_0}+\frac{\sigma c}{{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}(a-b+c)c=a+b$
As $c=a+b$
$\therefore V_A=\frac{\sigma }{{\epsilon }_0}\left(2a\right)$
$V_B=\frac{\sigma }{{\epsilon }_0}[\frac{a^2}{b}-b+c]=\frac{\sigma }{{\epsilon }_0}[\frac{a^2}{b}-b+a+b]$
$V_B=\frac{\sigma }{{\epsilon }_0}[\frac{a^2}{b}+a]=\frac{\sigma a}{{\epsilon }_0}\frac{(a+b)}{b}=\frac{{\sigma }_{0}ac}{{\epsilon }_{0}b}$
$V_C=\frac{\sigma }{{\epsilon }_0}[\frac{a^2}{b}-\frac{a^2}{c}+c]=\frac{\sigma a}{{\epsilon }_0}\left[\frac{a^2-b^2+c^2}{c}\right]=\frac{\sigma }{{\epsilon }_0}\left[\frac{(a+b)(a-b)+{(a+b)}^2}{c}\right]$
$V_C=\frac{\sigma }{{\epsilon }_0}\left[\frac{(a+b)}{c}(a-b+a+b)\right]=\frac{\sigma 2a}{{\epsilon }_0}$
we found that, $V_C$ = $V_A\ne V_B$