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(C) The electric field $E$ on the surface of a conducting sphere of radius $R$ and charge $q$ is given by $E = \frac{kq}{R^2}$.Given that the electric

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$(r_1 / r_2)$

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(C) The electric field $E$ on the surface of a conducting sphere of radius $R$ and charge $q$ is given by $E = \frac{kq}{R^2}$.Given that the electric fields near the surfaces are equal:$\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2}$$\frac{q_1}{q_2} = \frac{r_1^2}{r_2^2} \quad -(1)$The electric potential $V$ on the surface of a conducting sphere is given by $V = \frac{kq}{R}$.The ratio of their potentials is:$\frac{V_1}{V_2} = \frac{kq_1 / r_1}{kq_2 / r_2} = \frac{q_1}{q_2} \cdot \frac{r_2}{r_1}$Substituting the value of $\frac{q_1}{q_2}$ from equation $(1)$:$\frac{V_1}{V_2} = \left( \frac{r_1^2}{r_2^2} \right) \cdot \left( \frac{r_2}{r_1} \right) = \frac{r_1}{r_2}$Thus,the ratio of their electric potentials is $(r_1 / r_2)$.

Two conducting spheres of radii $r_1$ and $r_2$ have the same electric field intensity near their surfaces. The ratio of their electric potentials is

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