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(A) The surface charge density $\sigma$ is defined as the charge per unit area,$\sigma = \frac{q}{4 \pi R^2}$.Since the surface charge density is equa

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$R_1/R_2$

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(A) The surface charge density $\sigma$ is defined as the charge per unit area,$\sigma = \frac{q}{4 \pi R^2}$.Since the surface charge density is equal for both spheres,we have $\sigma_1 = \sigma_2 = \sigma$.Thus,the charge on the spheres is $q_1 = \sigma (4 \pi R_1^2)$ and $q_2 = \sigma (4 \pi R_2^2)$.The electric potential $V$ at the surface of a charged sphere is given by $V = \frac{kq}{R}$.Substituting the expression for $q$,we get $V = \frac{k (\sigma 4 \pi R^2)}{R} = k \sigma 4 \pi R$.Since $k$,$\sigma$,and $4 \pi$ are constants,the potential is directly proportional to the radius: $V \propto R$.Therefore,the ratio of their potentials is $\frac{V_1}{V_2} = \frac{R_1}{R_2}$.

Two charged spheres of radii $R_1$ and $R_2$ have equal surface charge density. The ratio of their potential is

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