Two isolated metallic solid spheres of radii | vedclass

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(D) Initial charges on the spheres are:$Q_1 = \sigma(4\pi R^2) = 4\pi R^2\sigma$$Q_2 = \sigma(4\pi(2R)^2) = 16\pi R^2\sigma$Total charge $Q = Q_1 + Q_

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$\frac{5}{6}$

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(D) Initial charges on the spheres are:$Q_1 = \sigma(4\pi R^2) = 4\pi R^2\sigma$$Q_2 = \sigma(4\pi(2R)^2) = 16\pi R^2\sigma$Total charge $Q = Q_1 + Q_2 = 20\pi R^2\sigma$.When connected by a wire,the potentials become equal:$V_1 = V_2 \implies \frac{kQ_1^{\prime}}{R} = \frac{kQ_2^{\prime}}{2R} \implies Q_2^{\prime} = 2Q_1^{\prime}$.Since charge is conserved,$Q_1^{\prime} + Q_2^{\prime} = 20\pi R^2\sigma$.Substituting $Q_1^{\prime} = \frac{Q_2^{\prime}}{2}$,we get $\frac{Q_2^{\prime}}{2} + Q_2^{\prime} = 20\pi R^2\sigma \implies \frac{3}{2}Q_2^{\prime} = 20\pi R^2\sigma \implies Q_2^{\prime} = \frac{40}{3}\pi R^2\sigma$.The new charge density of the bigger sphere is $\sigma^{\prime} = \frac{Q_2^{\prime}}{4\pi(2R)^2} = \frac{Q_2^{\prime}}{16\pi R^2}$.Substituting $Q_2^{\prime}$,we get $\sigma^{\prime} = \frac{40\pi R^2\sigma}{3 \cdot 16\pi R^2} = \frac{40\sigma}{48} = \frac{5}{6}\sigma$.Therefore,$\frac{\sigma^{\prime}}{\sigma} = \frac{5}{6}$.

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