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A charge Q is distributed over two concentric hollow spheres of radii r and $R (> r)$ such that their surface charge densities are equal. The potential at their common centre is

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

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

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

answer is D.

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Detailed Solution

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

$\frac{Q_{1}}{Q_{2}} = \frac{R^{2}}{r^{2}}$

And $Q_{1} + Q_{2} = Q$

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

$Q_{1} [\frac{R^{2} + r^{2}}{R^{2}}] = Q$

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

$Q_{2} = \frac{Q r^{2}}{R^{2} + r^{2}}$

$∴ V = \frac{K Q_{1}}{R} + \frac{K Q_{2}}{r}$

$= \frac{K Q R}{R^{2} + r^{2}} + \frac{K Q r}{R^{2} + r^{2}}$

$= \frac{K Q (R + r)}{R^{2} + r^{2}} = \frac{1}{4 π ε_{0}} \frac{Q (R + r)}{R^{2} + r^{2}}$

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