Two concentric, conducting, thin spherical shells A and B have radii ‘r’ and ‘R’ as shown in figure. Shell A carries a charge 'q' and B carries a charge ‘Q’. Shell A is supported on an insulating stand. The potential of shell A is

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

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

$\frac{1}{4 π \in_{0}} \frac{q}{r}$

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

answer is A.

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Detailed Solution

The potential at every point inside the outer shell B is $V_{1} = \frac{Q}{4 π \in_{0} R}$ , which is also the potential on the surface of inner shell A due to outer shell B. The potential on the surface of shell A due to its own charge is $V_{2} = \frac{q}{4 π \in_{0} r} .$ Therefore, the total potential of shell A is
$V_{A} = V_{1} + V_{2} = \frac{Q}{4 π \in_{0} R} + \frac{q}{4 π \in_{0} r}$
$= \frac{1}{4 π \in_{0}} (\frac{Q}{R} + \frac{q}{r})$