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)

$\frac {\sigma}{\epsilon_o}(R+r)$

Answer 2

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 $V =V_{1}+V_{2}$ $=\frac{q_{1}}{4 \pi \varepsilon_{0} R}+\frac{q_{2}}{4 \pi \varepsilon_{0} r}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{R}+\frac{q_{2}}{r}\right)$ $=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{R^{2}}{r^{2}} \frac{q_{2}}{R}\right)+\frac{q_{2}}{r}=\frac{q_{2}}{4 \pi \varepsilon_{0}}\left(\frac{R+r}{r^{2}}\right)$ $=\frac{q_{2}}{4 \pi r^{2} \varepsilon_{0}}(R+r)=\frac{\sigma}{\varepsilon_{0}}(R+r)$