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. $\frac{1}{4 \pi \left(\epsilon \right)_{0}} \, \frac{Q \left(R + r\right)}{R^{2} + r^{2}}$
• B. $\frac{1}{4 \pi \left(\epsilon \right)_{0}}\frac{Q \left(R - r\right)}{R^{2} + r^{2}}$
• C. $\frac{1}{4 \pi \left(\epsilon \right)_{0}}\frac{Q \left(R + r\right)^{2}}{R^{2} + r^{2}}$
• D. $\frac{1}{4 \pi \left(\epsilon \right)_{0}}\frac{Q \left(R + r\right)^{2}}{R^{2} + \left(2r\right)^{2}}$

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 \epsilon _{0}} \, \left[\frac{Q_{2}}{r} + \frac{Q_{R}}{R}\right]=\frac{1}{4 \pi \epsilon _{0}} \, . \, \frac{Q}{r^{2} + R^{2}} \, \left[r + R\right] \, $