Question

Two concentric spheres of radii $R$ and $r$ have positive charges $q_1$ and $q_2$ with equal surface charge densities. What is the electric potential at their common centre ?

• A. $\frac{\sigma }{{ϵ}_o}(R+r)$
• B. $\frac{\sigma }{{ϵ}_o}(R-r)$
• C. $\frac{\sigma }{{ϵ}_o}(\frac{1}{R}+\frac{1}{r})$
• D. $\frac{\sigma }{{ϵ}_o}(\frac{1}{r})$

Answer 1

Answer: (A)

Potential at the centre due to sphere with radius $R$,
$V_1=\frac{q_1}{4\pi {\epsilon }_{0}R}$
Potential at the centre due to sphere with radius $r$
$V_2=\frac{q_2}{4\pi {\epsilon }_{0}r}$
Given, surface charge densities for both spheres
$\sigma ={\sigma }_R={\sigma }_r$
i.e., $\frac{q_1}{4\pi {\epsilon }_0{R}^2}=\frac{q_2}{4\pi {\epsilon }_0{r}^2}$
$\Rightarrow a_1=\frac{R^2}{r^2}{q}_2$
Total potential at the centre due to both spheres
$V=V_1+V_2$
$=\frac{q_1}{4\pi {\epsilon }_{0}R}+\frac{q_2}{4\pi {\epsilon }_{0}r}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1}{R}+\frac{q_2}{r}\right)$
$=\frac{1}{4\pi {\epsilon }_0}\left(\frac{R^2}{r^2}\frac{q_2}{R}\right)+\frac{q_2}{r}=\frac{q_2}{4\pi {\epsilon }_0}\left(\frac{R+r}{r^2}\right)$
$=\frac{q_2}{4\pi r^2{\epsilon }_0}(R+r)=\frac{\sigma }{{\epsilon }_0}(R+r)$