A charge Q is distributed over two concentric conducting thin spherical shells radii r and R $(R > r)$ If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

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

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

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

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

answer is D.

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Detailed Solution

As the charge density is same the charge on the surface will be proportional to r²,

$Q_{r} = \frac{{Qr}^{2}}{r^{2} + R^{2}}; Q_{R} = \frac{{QR}^{2}}{r^{2} + R^{2}}$

$V = \frac{1}{4 {πε}_{0}} (\frac{Q_{r}}{r} + \frac{Q_{R}}{R})$ = $\frac{1}{4 π \in_{0}} \frac{(R + r)}{(R^{2} + r^{2})} Q$

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