Electrostatics Question 778
Question: A charge Q is distributed over two concentric hollow spheres of radii r and R(R>r) such that the surface densities are equal. The potential at the common center is $ \frac{1}{4\pi {\varepsilon _{0}}} $ times-
Options:
A) $ Q[ \frac{r+R}{r^{2}+R^{2}} ] $
B) $ \frac{Q}{2}[ \frac{r+R}{r^{2}+R^{2}} ] $
C) $ 2Q[ \frac{r+R}{r^{2}+R^{2}} ] $
D) zero
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ q _{r}+q _{R}=Q $ -.(1) $ \frac{q _{r}}{q _{R}}=\frac{4\pi r^{2}\sigma }{4\pi R^{2}\sigma }=\frac{r^{2}}{R^{2}}\Rightarrow \frac{q _{r}}{q _{R}}=\frac{r^{2}}{R^{2}} $ ..(2)
From (1) and (2) $ q _{r}=\frac{Qr^{2}}{R^{2}+r^{2}} $
and $ q _{R}=\frac{QR^{2}}{R^{2}+r^{2}} $
So, $ V=\frac{q _{r}}{4\pi {\varepsilon _{0}}r}+\frac{q _{R}}{4\pi {\varepsilon _{0}}R} $
$ \Rightarrow V=\frac{Q}{4\pi {\varepsilon _{0}}}( \frac{r+R}{r^{2}+R^{2}} ) $
