Question

Three concentric spherical metallic shells $\text{A, B}$ and $\text{C}$ of radii $a,b$ and $c$ $(a<b<c)$ have surface charge densities $\sigma ,-\sigma$ and $\sigma$ respectively. If the shells $\text{A}$ and $\text{C}$ are at the same potential, then the correct relation between $a,b$ and $c$ is-

• A. $a+c=b$
• B. $b+c=a$
• C. $a-b=c$
• D. $a+b=c$
  

Answer 1

The correct option is D $a+b=c$


Given, Three concentric spherical metallic shells $\text{A, B}$ and $\text{C}$ of radii $a,b$ and $c$ $(a<b<c)$ have surface charge densities $\sigma ,-\sigma$ and $\sigma$ respectively.




$\because \text{Surface charge density,}\sigma =\frac{q}{\text{Area}}=\frac{q}{4\pi r^2}$

$\Rightarrow q_a=\sigma \times 4\pi a^2{q}_b=\sigma \times 4\pi b^2{q}_c=\sigma \times 4\pi c^2$

Potential of shell $\text{A}$ is,

$V_A=\frac{1}{4\pi {\epsilon }_0}(\frac{4\pi a^2\sigma }{a}-\frac{4\pi b^2\sigma }{b}+\frac{4\pi c^2\sigma }{c})$

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

Potential of shell $\text{C}$ is,

$V_C=\frac{1}{4\pi {\epsilon }_0}(\frac{4\pi a^2\sigma }{c}-\frac{4\pi b^2\sigma }{c}+\frac{4\pi c^2\sigma }{c})$

$=\frac{\sigma }{{\epsilon }_0}(\frac{a^2}{c}-\frac{b^2}{c}+c)$

As given $V_A=V_C$

$\therefore \frac{\sigma }{{\epsilon }_0}(a-b+c)=\frac{\sigma }{{\epsilon }_0}(\frac{a^2}{c}-\frac{b^2}{c}+c)$

$a-b=\frac{(a-b)(a+b)}{c}$

$\therefore c=a+b$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.