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(D) Initial charges on the spheres are $q_1 = \sigma \cdot 4\pi R^2$ and $q_2 = \sigma \cdot 4\pi (3R)^2 = 9\sigma \cdot 4\pi R^2$.Total charge $Q = q

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$3$

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(D) Initial charges on the spheres are $q_1 = \sigma \cdot 4\pi R^2$ and $q_2 = \sigma \cdot 4\pi (3R)^2 = 9\sigma \cdot 4\pi R^2$.Total charge $Q = q_1 + q_2 = 10\sigma \cdot 4\pi R^2$.When the spheres are brought into contact, the total charge is redistributed such that they reach the same potential $V$.$V = \frac{k q_1'}{R} = \frac{k q_2'}{3R}$, where $q_1'$ and $q_2'$ are the new charges.This implies $q_2' = 3q_1'$.Since $q_1' + q_2' = Q$, we have $4q_1' = 10\sigma \cdot 4\pi R^2$, so $q_1' = 2.5\sigma \cdot 4\pi R^2$ and $q_2' = 7.5\sigma \cdot 4\pi R^2$.The new surface charge densities are $\sigma_1 = \frac{q_1'}{4\pi R^2} = 2.5\sigma$ and $\sigma_2 = \frac{q_2'}{4\pi (3R)^2} = \frac{7.5\sigma}{9} = \frac{2.5\sigma}{3}$.Therefore, the ratio $\frac{\sigma_1}{\sigma_2} = \frac{2.5\sigma}{2.5\sigma / 3} = 3$.

Two metal spheres of radius $R$ and $3R$ have the same surface charge density $\sigma$. If they are brought in contact and then separated, the surface charge density on the smaller and bigger sphere becomes $\sigma_1$ and $\sigma_2$, respectively. The ratio $\frac{\sigma_1}{\sigma_2}$ is:

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