Two spherical conductors A and B of radii 1 | vedclass

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(C) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_1 = V_2$.Since the potential of a

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$2 : 1$

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(C) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_1 = V_2$.Since the potential of a spherical conductor is given by $V = K \frac{Q}{r}$,we have $K \frac{Q_1}{r_1} = K \frac{Q_2}{r_2}$,which implies $\frac{Q_1}{r_1} = \frac{Q_2}{r_2}$ or $\frac{Q_1}{Q_2} = \frac{r_1}{r_2}$.The electric field at the surface of a spherical conductor is $E = K \frac{Q}{r^2}$.Therefore,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{K Q_1 / r_1^2}{K Q_2 / r_2^2} = \frac{Q_1}{Q_2} 	imes \frac{r_2^2}{r_1^2}$.Substituting $\frac{Q_1}{Q_2} = \frac{r_1}{r_2}$,we get $\frac{E_1}{E_2} = \frac{r_1}{r_2} 	imes \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$.Given $r_1 = 1 \ mm$ and $r_2 = 2 \ mm$,the ratio is $\frac{E_1}{E_2} = \frac{2}{1} = 2 : 1$.

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