A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R$ $>r$ )…

A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R$ $>r$ ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is
  1. $\frac{1}{4 \pi \varepsilon_0} \frac{(R-r) Q}{\left(R^2+r^2\right)}$
  2. $\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{2\left(R^2+r^2\right)}$
  3. $\left.\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{\left(R^2+r^2\right.}\right)$
  4. $\frac{1}{4 \pi \varepsilon_0} \frac{(R-r) Q}{2\left(R^2+r^2\right)}$

Solution

Let $q_1$ and $q_2$ be charge on two spheres of radius ' $r$ ' and ' $R$ ' respectively As, $q_1+q_2=\mathrm{Q}$ and $\sigma_1=\sigma_2 \quad$ [Surface charge density are equal] $ \therefore \frac{q_1}{r \pi r^2}=\frac{q_2}{4 \pi R^2} $ So, $q_1=\frac{Q r^2}{R^2+r^2}$ and $q_2=\frac{Q R^2}{R^2+r^2}$ Now, potential, $V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r}+\frac{q_2}{R}\right]$ $ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q r}{R^2+r^2}+\frac{Q R}{R^2+r^2}\right] \\ & =\frac{Q(R+r)}{R^2+r^2} \frac{1}{4 \pi \varepsilon_0} \end{aligned} $

Asked in: JEE Main 2012 (19 May Online)

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