Two isolated metallic spheres of radii R and | vedclass

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(D) Initial charges on the spheres are:$q_1 = \sigma \cdot 4\pi R^2$$q_2 = \sigma \cdot 4\pi (2R)^2 = 4\sigma \cdot 4\pi R^2 = 4q_1$Total charge $Q =

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$5:6$

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(D) Initial charges on the spheres are:$q_1 = \sigma \cdot 4\pi R^2$$q_2 = \sigma \cdot 4\pi (2R)^2 = 4\sigma \cdot 4\pi R^2 = 4q_1$Total charge $Q = q_1 + q_2 = 5q_1 = 5\sigma \cdot 4\pi R^2$.When connected, the spheres reach a common potential $V = \frac{kQ_1}{R} = \frac{kQ_2}{2R}$, where $Q_1$ and $Q_2$ are new charges.Thus, $Q_2 = 2Q_1$. Since $Q_1 + Q_2 = Q$, we have $3Q_1 = 5q_1 \implies Q_1 = \frac{5}{3}q_1$ and $Q_2 = \frac{10}{3}q_1$.The new charge density of the larger sphere is $\sigma_1 = \frac{Q_2}{4\pi(2R)^2} = \frac{10/3 \cdot \sigma \cdot 4\pi R^2}{16\pi R^2} = \frac{10}{3 \cdot 16} \sigma = \frac{10}{48} \sigma = \frac{5}{24} \sigma$.Wait, re-evaluating: $Q_2 = \frac{10}{3} \sigma (4\pi R^2)$. $\sigma_1 = \frac{Q_2}{16\pi R^2} = \frac{10/3 \cdot \sigma \cdot 4\pi R^2}{16\pi R^2} = \frac{10}{3 \cdot 4} \sigma = \frac{5}{6} \sigma$. Thus, the ratio $\sigma_1/\sigma = 5/6$.

Two isolated metallic spheres of radii $R$ and $2R$ are charged such that both have the same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the larger sphere is $\sigma_1$. The ratio $\sigma_1$ to $\sigma$ is

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