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. Thus,$V_A = V_B$.Using the formula for po

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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. Thus,$V_A = V_B$.Using the formula for potential $V = \frac{KQ}{r}$,we have:$\frac{KQ_A}{r_A} = \frac{KQ_B}{r_B}$$\Rightarrow \frac{Q_A}{r_A} = \frac{Q_B}{r_B} \Rightarrow \frac{Q_A}{Q_B} = \frac{r_A}{r_B}$The electric field at the surface of a spherical conductor is given by $E = \frac{KQ}{r^2}$.Therefore,the ratio of the electric fields $E_A$ and $E_B$ is:$\frac{E_A}{E_B} = \frac{K Q_A / r_A^2}{K Q_B / r_B^2} = \frac{Q_A}{Q_B} 	imes \left( \frac{r_B}{r_A} ight)^2$Substituting $\frac{Q_A}{Q_B} = \frac{r_A}{r_B}$ into the equation:$\frac{E_A}{E_B} = \left( \frac{r_A}{r_B} ight) 	imes \left( \frac{r_B}{r_A} ight)^2 = \frac{r_B}{r_A}$Given $r_A = 1 \ mm$ and $r_B = 2 \ mm$:$\frac{E_A}{E_B} = \frac{2 \ mm}{1 \ mm} = \frac{2}{1} = 2 : 1$Since the distance between the spheres is large compared to their diameters,the induced effects may be ignored.

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