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(B) When two charged spheres are joined by a wire,they reach the same electric potential $V$.For a sphere of radius $R$ and charge $Q$,the potential i

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$b/a$

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(B) When two charged spheres are joined by a wire,they reach the same electric potential $V$.For a sphere of radius $R$ and charge $Q$,the potential is $V = \frac{kQ}{R}$.Since $V_1 = V_2$,we have $\frac{kQ_1}{a} = \frac{kQ_2}{b}$,which implies $\frac{Q_1}{Q_2} = \frac{a}{b}$.The electric field at the surface of a sphere is given by $E = \frac{kQ}{R^2}$.Therefore,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{kQ_1/a^2}{kQ_2/b^2} = \frac{Q_1}{Q_2} 	imes \frac{b^2}{a^2}$.Substituting $\frac{Q_1}{Q_2} = \frac{a}{b}$,we get $\frac{E_1}{E_2} = \frac{a}{b} 	imes \frac{b^2}{a^2} = \frac{b}{a}$.

Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of the electric field at the surfaces of the spheres is

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