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(D) Let the charges on the spheres of radii $R$ and $r$ be $Q_1$ and $Q_2$ respectively,such that $Q_1 + Q_2 = Q$.Given that the surface charge densit

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the charges on the spheres of radii $R$ and $r$ be $Q_1$ and $Q_2$ respectively,such that $Q_1 + Q_2 = Q$.Given that the surface charge densities are equal,$\sigma = \frac{Q_1}{4\pi R^2} = \frac{Q_2}{4\pi r^2}$.This implies $\frac{Q_1}{R^2} = \frac{Q_2}{r^2} = k'$,where $k'$ is a constant.Thus,$Q_1 = k' R^2$ and $Q_2 = k' r^2$.Since $Q_1 + Q_2 = Q$,we have $k'(R^2 + r^2) = Q$,so $k' = \frac{Q}{R^2 + r^2}$.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 $Q_2$:$V = \frac{1}{4\pi\varepsilon_0} \left( \frac{k' R^2}{R} + \frac{k' r^2}{r} \right) = \frac{k'}{4\pi\varepsilon_0} (R + r)$.Substituting $k' = \frac{Q}{R^2 + r^2}$:$V = \frac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)}$.

If on two concentric hollow spheres of radii $R$ and $r$ $(R < r)$,a total charge $Q$ is distributed such that their surface charge densities are equal,then the potential at their common centre is:

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