Two charged conducting spheres S_1 and S_2 of | vedclass

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(D) When two conductors are connected by a wire, their potentials become equal $(V_1 = V_2)$.For a conducting sphere of radius $R$ and charge $q$, the

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$\frac{9}{4}$

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(D) When two conductors are connected by a wire, their potentials become equal $(V_1 = V_2)$.For a conducting sphere of radius $R$ and charge $q$, the potential is $V = \frac{kq}{R}$.Since the potentials are equal, we have $\frac{kq_1}{R_1} = \frac{kq_2}{R_2}$, which implies $\frac{q_1}{q_2} = \frac{R_1}{R_2}$.The electric field on the surface of a sphere is $E = \frac{kq}{R^2}$.Therefore, the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{kq_1/R_1^2}{kq_2/R_2^2} = \frac{q_1}{q_2} \cdot \frac{R_2^2}{R_1^2}$.Substituting $\frac{q_1}{q_2} = \frac{R_1}{R_2}$ into the equation, we get $\frac{E_1}{E_2} = \frac{R_1}{R_2} \cdot \frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}$.Given $R_1 = 8 	ext{ cm}$ and $R_2 = 18 	ext{ cm}$, the ratio is $\frac{E_1}{E_2} = \frac{18}{8} = \frac{9}{4}$.

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