Electrostatics Question 231

Question: If on the concentric hollow spheres of radii $ r $ and $ R(>r) $ the charge $ Q $ is distributed such that their surface densities are same then the potential at their common centre is [IIT 1981; MP PMT 2003]

Options:

A) $ \frac{Q(R^{2}+r^{2})}{4\pi {\varepsilon _{0}}(R+r)} $

B) $ \frac{QR}{R+r} $

C) Zero

D) $ \frac{Q(R+r)}{4\pi {\varepsilon _{0}}(R^{2}+r^{2})} $

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Answer:

Correct Answer: D

Solution:

$ q _{1}+q _{2}=Q $

and $ \frac{q _{1}}{4\pi r^{2}}=\frac{q _{2}}{4\pi R^{2}} $ (given) $ q _{1}=\frac{Qr^{2}}{R^{2}+r^{2}} $

and $ q _{2}=\frac{QR^{2}}{R^{2}+r^{2}} $

Potential at common centre $ \frac{1}{4\pi {\varepsilon _{0}}}[ \frac{Qr^{2}}{(R^{2}+r^{2})r}+\frac{QR^{2}}{(R^{2}+r^{2})R} ]=\frac{Q(R+r)}{4\pi {\varepsilon _{0}}(R^{2}+r^{2})} $



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