A charge Q is distributed over two concentric hollow, spheres of radius r and R(<r) such that the surface densities are equal. The potential at the common centre is :

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$\frac{Q}{R + r}$

$\frac{Q (R + r)}{4 π ε_{0} (R^{2} + r^{2})}$

$\frac{Q (R^{2} + r^{2})}{4 π ε_{0} (R + r)}$

Zero

answer is B.

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Detailed Solution

since, the surface densities are equal, hence

$\frac{q_{1}}{4 π r^{2}} = \frac{q_{2}}{4 π R^{2}}$

(where $q_{1} + q_{2} = Q$ )

or $\frac{q_{1}}{r^{2}} = \frac{q_{2}}{R^{2}} = \frac{q_{1} + q_{2}}{r^{2} + R^{2}} = \frac{Q}{r^{2} + R^{2}}$

$\begin{array}{l} ∴ q_{1} = \frac{Q}{r^{2} + R^{2}} \times r^{2} \\ and q_{2} = \frac{Q}{r^{2} + R^{2}} \times R^{2} \end{array}$

So, potential at the common centre,

$\begin{array}{l} V = \frac{q_{1}}{4 π ε_{0} r} + \frac{q_{2}}{4 π ε_{0} R} = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r} + \frac{q_{2}}{R}) \\ = \frac{1}{4 π ε_{0}} [\frac{Q}{R^{2} + r^{2}} \times \frac{r^{2}}{r} + \frac{Q}{R^{2} + r^{2}} \times \frac{R^{2}}{R}] \\ = \frac{1}{4 π ε_{0}} \frac{Q (R + r)}{(R^{2} + r^{2})} \end{array}$