Question

Three concentric charged metallic spherical shells $A,B$ and $C$ have radii $a,b$ and $c;$ charge densities $\sigma ,-\sigma$ and $\sigma$ and potentials $V_A,V_B$ and $V_C$ respectively. Then which of the following relations is correct?

• A. $V_A=(a+b+c)\frac{\sigma }{{\epsilon }_0}$
• B. $V_B=\left(\frac{a^2}{b}-b+c\right)\frac{\sigma }{{\epsilon }_0}$
• C. $V_C=\left(\frac{a^2+b^2}{c}+c\right)\frac{\sigma }{{\epsilon }_0}$
• D. $V_A=V_B=V_C=(a+b+c)\frac{\sigma }{{\epsilon }_0}$

Answer 1

Answer: (B)

Potential due to charged shell at an outside point
$=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
where $r$ is the distance of point from the centre of the shell while that inside the shell $=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$
where $R$ is the radius of the shell
Charges on the three shells are given by
$q_A=4\pi a^2\sigma$,
$q_B=-4\pi b^2\sigma$,
$q_c=4\pi c^2\sigma$
$\therefore V_A=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{4\pi a^2\sigma }{a}-\frac{4\pi b^2\sigma }{b}+\frac{4\pi c^2\sigma }{c}\right]=\frac{\sigma }{{\epsilon }_0}(a-b+c)$
$V_B=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_A}{b}+\frac{q_B}{b}+\frac{q_C}{c}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{4\pi a^2\sigma }{b}-\frac{4\pi b^2\sigma }{b}+\frac{4\pi c^2\sigma }{c}\right]$
$=\frac{4\pi \sigma }{4\pi {\epsilon }_0}\left[\frac{a^2}{b}-b+c\right]=\frac{\sigma }{{\epsilon }_0}\left(\frac{a^2}{b}-b+c\right)$
$V_C=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_A}{c}+\frac{q_B}{c}+\frac{q_C}{c}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{4\pi a^2\sigma }{c}-\frac{4\pi b^2\sigma }{c}+\frac{4\pi c^2\sigma }{c}\right]$
$=\frac{\sigma }{{\epsilon }_0}\left(\frac{a^2-b^2}{c}+c\right)$