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

Answer 1

Answer: (A)

Potential at the centre due to sphere with radius $R$,
$V_{1}=\frac{q_{1}}{4 \pi \varepsilon_{0} R}$
Potential at the centre due to sphere with radius $r$
$V_{2}=\frac{q_{2}}{4 \pi \varepsilon_{0} \,r}$
Given, surface charge densities for both spheres
$\sigma=\sigma_{R}=\sigma_{r}$
i.e., $ \frac{q_{1}}{4 \pi \varepsilon_{0} R^{2}}=\frac{q_{2}}{4 \pi \varepsilon_{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 \varepsilon_{0} R}+\frac{q_{2}}{4 \pi \varepsilon_{0} r}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{R}+\frac{q_{2}}{r}\right) $
$=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{R^{2}}{r^{2}} \frac{q_{2}}{R}\right)+\frac{q_{2}}{r}=\frac{q_{2}}{4 \pi \varepsilon_{0}}\left(\frac{R+r}{r^{2}}\right) $
$=\frac{q_{2}}{4 \pi r^{2} \varepsilon_{0}}(R+r)=\frac{\sigma}{\varepsilon_{0}}(R+r)$