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})} $
Show Answer
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})} $
