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

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 that $q_1 + q_2 = Q$.Since surface charge densities are equal,$\sigma = \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}$.This implies $\frac{q_1}{r^2} = \frac{q_2}{R^2}$,so $q_1 = q_2 \frac{r^2}{R^2}$.Substituting this into the total charge equation: $q_2 \frac{r^2}{R^2} + q_2 = Q \implies q_2 \left( \frac{r^2 + R^2}{R^2} \right) = Q \implies q_2 = \frac{Q R^2}{R^2 + r^2}$.Similarly,$q_1 = \frac{Q r^2}{R^2 + r^2}$.The potential at the common centre is the sum of potentials due to both spheres: $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.Substituting the values of $q_1$ and $q_2$: $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right)$.$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r}{R^2 + r^2} + \frac{Q R}{R^2 + r^2} \right) = \frac{Q(R + r)}{4\pi \varepsilon_0(R^2 + r^2)}$.

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

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