Question

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

• A. $\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{Q\left(R+r\right)}{R^2+r^2}$
• B. $\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{Q\left(R-r\right)}{R^2+r^2}$
• C. $\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{Q{\left(R+r\right)}^2}{R^2+r^2}$
• D. $\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{Q{\left(R+r\right)}^2}{R^2+{\left(2r\right)}^2}$

Answer 1

Answer: (A)

Since surface charge densities are equal, Q is divided in the ratio of surface areas (i.e., $r^2.R^2$ )
$Q_r=\frac{r^{2}Q}{r^2+R^2}{Q}_R=\frac{R^{2}Q}{r^2+R^2}$
$\therefore V=\frac{1}{4\pi {ϵ}_0}\left[\frac{Q_2}{r}+\frac{Q_R}{R}\right]=\frac{1}{4\pi {ϵ}_0}.\frac{Q}{r^2+R^2}\left[r+R\right]$