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(D) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively. Given $q_1 + q_2 = Q$.Since surface charge densities are equa

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$\frac{Q(R + r)}{4\pi\varepsilon_0(R^2 + r^2)}$

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(D) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively. Given $q_1 + q_2 = Q$.Since surface charge densities are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.Using the property of ratios,$\frac{q_1}{r^2} = \frac{q_2}{R^2} = \frac{q_1 + q_2}{r^2 + R^2} = \frac{Q}{r^2 + R^2}$.Thus,$q_1 = \frac{Q r^2}{r^2 + R^2}$ and $q_2 = \frac{Q R^2}{r^2 + R^2}$.The potential at the common centre is $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.Substituting the values of $q_1$ and $q_2$,we get $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q r^2}{r(r^2 + R^2)} + \frac{Q R^2}{R(r^2 + R^2)} \right)$.$V = \frac{Q}{4\pi\varepsilon_0(r^2 + R^2)} (r + R)$.

$A$ charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ such that their surface charge densities are equal. The potential at the common centre is

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