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(D) Let $q_1$ and $q_2$ be the charges on the inner and outer shells respectively. Given $Q = q_1 + q_2$ ... $(i)$Since surface charge densities are e

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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 inner and outer shells respectively. Given $Q = q_1 + q_2$ ... $(i)$Since surface charge densities are equal,$\sigma_1 = \sigma_2$,so $\frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \implies \frac{q_1}{q_2} = \frac{r^2}{R^2}$ ... $(ii)$From $(i)$ and $(ii)$,we get $q_1 = \frac{Q r^2}{R^2 + r^2}$ and $q_2 = \frac{Q R^2}{R^2 + r^2}$.The potential at the center is the sum of potentials due to both shells: $V = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q_1}{r} + \frac{q_2}{R} \right]$.Substituting the values of $q_1$ and $q_2$: $V = \frac{1}{4\pi \varepsilon_0} \left[ \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right] = \frac{1}{4\pi \varepsilon_0} \left[ \frac{Qr}{R^2 + r^2} + \frac{QR}{R^2 + r^2} \right]$.Thus,$V = \frac{Q(R + r)}{4\pi \varepsilon_0(R^2 + r^2)}$.

$A$ total charge $Q$ is distributed between two concentric spherical shells of radii $r$ and $R$ $(R > r)$ such that their surface charge densities are equal. What is the electric potential at their common center?

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