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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 $\sigma$

Vedclass · Accepted Answer

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

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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 $\sigma$ are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.This implies $\frac{q_1}{q_2} = \frac{r^2}{R^2}$.Using the ratio,we find $q_1 = \frac{Qr^2}{R^2 + r^2}$ and $q_2 = \frac{QR^2}{R^2 + r^2}$.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 $V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Qr}{R^2 + r^2} + \frac{QR}{R^2 + r^2} \right) = \frac{Q(R + r)}{4\pi\varepsilon_0(R^2 + r^2)}$.

If on two concentric hollow spheres of radii $r$ and $R$ $(R > r)$,the 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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