Question

Three concentric spherical shells have radii $a,b$ and $c$ $(a<b<c)$ and have surface charge densities $\sigma ,-\sigma \text{and}\sigma$ respectively. If $V_A,V_B\text{and}{V}_C$ denote the potentials of the three shells, then for $c=a+b$. We have

• A. $V_C=V_B=V_A$
• B. $V_C=V_A\ne V_B$
• C. $V_C=V_B\ne V_A$
• D. $V_C\ne V_B\ne V_A$

Answer 1

The correct option is B $V_C=V_A\ne V_B$

Potential of shell A is
$V_A=\frac{\sigma a}{{\epsilon }_0}+\frac{(-\sigma )b}{{\epsilon }_0}+\frac{\sigma \left(c\right)}{{\epsilon }_0}$
$\Rightarrow V_A=\frac{\sigma }{{\epsilon }_0}(a-b+c)$
$\Rightarrow V_A=\frac{\sigma }{{\epsilon }_0}\left(2a\right)$ $\because c=a+b$

Potential of shell B is
$V_B=\frac{\sigma }{{\epsilon }_0}(\frac{a^2}{b}-b+c)$
$\Rightarrow V_B=\frac{\sigma }{{\epsilon }_0}(\frac{a^2-b^2}{b}+c)$
$V_B=\frac{\sigma }{{\epsilon }_0}(\frac{(a-b)(a+b)}{b}+c)$
$V_B=\frac{\sigma }{{\epsilon }_0}(\frac{(a-b)c}{b}+c)=\frac{\sigma }{{\epsilon }_0}\left(\frac{ac}{b}\right)$

Potential of shell C is
$V_C=\frac{\sigma }{{\epsilon }_0}(\frac{a^2}{c}-\frac{b^2}{c}+c)$
$\Rightarrow V_C=\frac{\sigma }{{\epsilon }_0}(\frac{a^2-b^2}{c}+c)$
$\Rightarrow V_C=\frac{\sigma }{{\epsilon }_0}(a-b+c)[\because c=a+b]$
$\therefore V_C=\frac{\sigma }{{\epsilon }_0}\left(2a\right)$
$\Rightarrow V_C=V_A\ne V_B$