Question

Two concentric hollow spherical shells have radii $r$ and $R$ $(R>>r)$. A charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is:

• A. $\frac{Q(R+r)}{4\pi {ϵ}_0(R^2+r^2)}$
• B. $\frac{Q(R^2+r^2)}{4\pi {ϵ}_0(R+r)}$
• C. $\frac{Q}{R+r}$
• D. $0$

Answer 1

The correct option is A $\frac{Q(R+r)}{4\pi {ϵ}_0(R^2+r^2)}$

If $q_1$ and $q_2$ are the charges on spheres of radii $r$ and $R$ respectively, then $q_1+q_2=Q$

According to the given situation:

$\Rightarrow {\sigma }_1={\sigma }_2$

$\Rightarrow \frac{q_1}{4\pi r^2}=\frac{q_2}{4\pi R^2}$

$\Rightarrow \frac{q_1}{q_2}=\frac{r^2}{R^2}$

$\Rightarrow q_1=\frac{Q{r}^2}{r^2+R^2}$ and $q_2=\frac{Q{R}^2}{r^2+R^2}$

Now as potential inside a conducting sphere is equal to its surface, so potential at the common centre:

$V=V_1+V_2=\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{r}+\frac{q_2}{R}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{Qr}{R^2+r^2}+\frac{QR}{R^2+r^2}]$

$=\frac{1}{4\pi {ϵ}_0}\frac{Q(R+r)}{R^2+r^2}$