Question

Three concentric metallic shells $A$, $B$ and $C$ of radii $a,b$ and $c(a<b<c)$ have surface charge densities, $\sigma ,-\sigma$ and $\sigma$ respectively. It is found that no work is required to bring a charge $q$ from shell $A$ to shell $C$ then obtain the relation between the radii $a,b$ and $c$ :

• A. $c=a+b$
• B. $b=\frac{a+c}{2}$
• C. $b=\sqrt{ac}$
• D. $c=ab$

Answer 1

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

The potential at a distance $r$ due to a uniformly charged spherical shell of radius $R$ is given by

$V=\frac{Q}{4\pi {ϵ}_{0}R}$ for $r<R$ and $V=\frac{Q}{4\pi {ϵ}_{0}r}$ for $r>R$

Here, $Q=4\pi R^2\sigma$

$A$ is the interior of all shells.

Thus, the potential at $A$ due to the charges is $V_A=V_{A,A}+V_{A,B}+V_{A,C}=\frac{\sigma }{{ϵ}_0}(a-b+c)$

$C$ is outside of the all shells.

Thus, the potential at $C$ due to the charges is $V_C=V_{C,A}+V_{C,B}+V_{C,C}=\frac{\sigma }{{ϵ}_{0}c}(a^2-b^2+c^2)$

Since no work is done for moving the charge between $A$ and $C$, $V_A=V_C$

Thus, $(a-b+c)c=a^2-b^2+c^2\Rightarrow (a-b)c=a^2-b^2\Rightarrow c=a+b$