Question

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

• A. $a=b+c$
• B. $c=a+b$
• C. $b=a+c$
• D. $2a=b-c$

Answer 1

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

Let us consider the general case of a metallic shell in which we consider three points x, y and z which lies inside the surface, on the surface and outside the surface respectively.
So we, know in case of metallic sphere,
potential inside the sphere = potential on the surface of sphere
$⟹V_x=V_y$
So, on shell A the potential due shell B and C will be same as that on their surfaces
$⟹V_A=\frac{k\sigma 4\pi a^2}{a}-\frac{k\sigma 4\pi b^2}{b}+\frac{k\sigma 4\pi c^2}{c}$ (we know Charge = $\sigma \times 4\pi r^2$)
$⟹V_A=\frac{\sigma }{{ϵ}_0}[a-b+c]$
IIy $V_B=\frac{k\sigma 4\pi a^2}{b}-\frac{k\sigma 4\pi b^2}{b}+\frac{k\sigma 4\pi c^2}{c}$
$⟹V_B=\frac{\sigma }{{ϵ}_0}[\frac{a^2}{b}-b+c]$
$V_C=\frac{k\sigma 4\pi a^2}{c}-\frac{k\sigma 4\pi b^2}{c}+\frac{k\sigma 4\pi c^2}{c}$
$⟹V_C=\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2+c^2}{c}\right]$
As $V_A=V_C$
$⟹\frac{\sigma }{{ϵ}_0}[a-b+c]$ = $\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2+c^2}{c}\right]$
$⟹c=a+b$