Question

A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface charge densities are equal. Find the potential at the common centre,

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

Answer 1

The correct option is B $\frac{Q}{4\pi {\epsilon }_0}\left(\frac{R+r}{R^2+r^2}\right)$

Let the charges $q_1$ and $q_2$ are distributed on hollow spheres of radii r and R respectively and common charge density be $\sigma$ then
$\sigma =\frac{q_1}{4\pi r^2}=\frac{q_2}{4\pi R^2}$
This means $q_1=\frac{r^2}{R^2}{q}_2$
Also, $q_1+q_2=Q$

From above two equations:
$q_1=\frac{Q{r}^2}{r^2+R^2}$ and $q_2=\frac{Q{R}^2}{r^2+R^2}$
Now, potential at the common center will be

$V=\frac{q_1}{4\pi {ϵ}_{0}r}+\frac{q_2}{4\pi {ϵ}_{0}R}$
After putting values of $q_1$ and $q_2$
$V=\frac{Q}{4\pi {ϵ}_0}\left(\frac{r+R}{R^2+r^2}\right)$