Question

A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R>r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

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

Answer 1

Answer: (C)

Let $q_1$ and $q_2$ be charge on two spheres of radius ${}^{\prime }{r}^{\prime }$ and ${}^{\prime }{R}^{\prime }$ respectively
As, $q_1+q_2=Q$
and ${\sigma }_1+{\sigma }_2$ [Surface charge density are equal]
$\therefore \frac{q_1}{r\pi r^2}=\frac{q_2}{4\pi R^2}$
So, $q_1=\frac{Q{r}^2}{R^2+r^2}$ and $q_2=\frac{Q{R}^2}{R^2+r^2}$
Now, potential, $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{r}+\frac{q_2}{R}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{Qr}{R^2+r^2}+\frac{Qr}{R^2+r^2}\right]$
$=\frac{Q\left(R+r\right)}{R^2+r^2}\frac{1}{4\pi {\epsilon }_0}$