Question

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

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

Answer 1

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

The concentric spherical shells have radii $=a,b$ and $c(a<b<c)$ and have surface charge densities $=+6,-6,+6$ respectively.

If $V_A$, $V_B$, $V_C$ denote the potentials of three shells then for $c=a+b$ we have

$V=$ potential for a concentric shell $\left[\frac{1}{4\pi {ϵ}_0}\right]\frac{q}{r}⟶\left(1\right)$

$\sigma =$ surface charge density $=\frac{q}{A}=\frac{q}{4\pi {ϵ}_0}\frac{q}{4\pi r^2}⟶\left(2\right)$

from $\left(1\right)$, $V_A=\frac{1}{4\pi {ϵ}_0}[\left(\frac{q_A}{a}\right)+\frac{q_B}{b}+\frac{q_C}{c}]$

from $\left(2\right)$, $q=6$ $4\pi r^2$ also,

${\sigma }_A=\sigma$

${\sigma }_B=-\sigma$

${\sigma }_C=\sigma$

$\therefore$ $V_A=\frac{1}{4\pi {ϵ}_0}[\frac{4\pi {\sigma a}^2}{a}-\frac{4\pi {\sigma b}^2}{b}+\frac{4\pi \sigma c^2}{c}]⟶\left(3\right)$

$V_A=\frac{1}{{ϵ}_0}[{\sigma }_a-{\sigma }_b+{\sigma }_c]$

$=\frac{\sigma }{{ϵ}_0}(a-b+c)⟶\left(3\right)$

Similarly,

$V_B=\left[\frac{1}{4\pi {ϵ}_0}\right][\frac{q_A}{a}+\frac{q_B}{b}+\frac{q_C}{c}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{4\pi {\sigma a}^2}{a}-\frac{4\pi {\sigma b}^2}{b}+\frac{4\pi \sigma c^2}{c}]$

$V_B=\frac{\sigma }{{ϵ}_0}[\frac{a^2}{b}-b+c]⟶\left(4\right)$

$V_C=\frac{\sigma }{{ϵ}_0}[\frac{a^2}{c}-\frac{b^2}{c}+c]⟶\left(5\right)$

Putting $c=a+b$

$V_A=V_C>V_B$

i.e $V_A=V_C\ne V_B$