Question

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

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

Answer 1

Answer: (D)

$q_A=4\pi a^2\sigma$
$q_B=-4\pi b^2\sigma$
$q_C=4\pi c^2\sigma ,c=a+b$
$V_A=\frac{1}{4\pi {\in }_0}\left(\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}\right)$
$=\frac{2\sigma a}{{ϵ}_0}$
$V_B=\frac{1}{4\pi {\in }_0}\left(\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}\right)$
$=\frac{\sigma }{{ϵ}_0}\left(\frac{a^2}{b}-b+c\right)$
$=\frac{\sigma }{{ϵ}_0}\left(a+\frac{a^2}{b}\right)$
$V_C=\frac{1}{4\pi {\in }_0}\left(\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}\right)$
$=\frac{\sigma }{{ϵ}_0}\left(\frac{a^2-b}{c}+c\right)=\frac{2\sigma a}{{ϵ}_0}$
So, $V_C=V_A\ne V_B$