Question

A charge Q is uniformly distributed over the surface of two concentric conducting spheres of radii R and r $(R>r)$ such that surface charge densities are same for both spheres. Then potential at the common center of these spheres is

• A. $\frac{kQ(R+r)}{Rr}$
• B. $\frac{kQ(R+r)}{(R^2+r^2)}$
• C. $\frac{kQ}{\sqrt{R^2+r^2}}$
• D. $kQ(\frac{1}{R}-\frac{1}{r})$

Answer 1

The correct option is B $\frac{kQ(R+r)}{(R^2+r^2)}$

charge Q is uniformly distributed over surface of the concentric conducting spheres of radii$(R>r)R\&r.$

surface change density are same for both the sphere.

suppose $q_1$and$q_2$ be charges on 2 concentric spheres of radii $r\&R$ respectively.

since surface charge densities are same;

so,

$=\frac{q_1}{4\pi r^2}=\frac{q^2}{4\pi r^2}$

or$q_1=\frac{q_2{r}_2}{R^2}⟶\left(1\right)$

also given$,q_1+q_2=Q$

from $1\left)\&2\right);$we get

$q_2=\frac{Q}{1+\frac{r^2}{R^2}}=\frac{Q{R}^2}{r^2+R^2}$

from$\left(2\right);q_1=Q-q_2=Q-\frac{Q{R}^2}{r^2+R^2}=\frac{Q{R}^2}{r^2+R^2}$

using superposition principle, potential at common center is :

$v=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{r}+\frac{q_2}{R}]$

substituting value of $q_1\&q_2,$we get

$=\frac{1}{4\pi {\epsilon }_0}[\frac{Qr}{r^2+R^2}+\frac{Qr}{r^2+R^2}]=\frac{1}{4\pi {\epsilon }_0}\frac{Q[r+R]}{r^2+R^2}=\frac{KQ[r+R]}{r^2+R^2}wherek=\frac{1}{4\pi {\epsilon }_0}$

so the answer is option (b).