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(C) When two conductors are connected by a thin conducting wire,charge flows until they reach the same potential $V$.Since $V_1 = V_2$,we have $\frac{

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$r: R$

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(C) When two conductors are connected by a thin conducting wire,charge flows until they reach the same potential $V$.Since $V_1 = V_2$,we have $\frac{k q_1}{R} = \frac{k q_2}{r}$,which implies $\frac{q_1}{q_2} = \frac{R}{r}$.The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi R^2}$.Therefore,the ratio of the surface charge densities is:$\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi R^2)}{q_2 / (4 \pi r^2)} = \left( \frac{q_1}{q_2} \right) \left( \frac{r^2}{R^2} \right)$.Substituting $\frac{q_1}{q_2} = \frac{R}{r}$,we get:$\frac{\sigma_1}{\sigma_2} = \left( \frac{R}{r} \right) \left( \frac{r^2}{R^2} \right) = \frac{r}{R}$.Thus,the ratio is $r: R$.

$A$ charged spherical conductor of radius $R$ is connected momentarily to another uncharged spherical conductor of radius $r$ by means of a thin conducting wire. Then,the ratio of the surface charge density of the first to the second conductor is:

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