Two uniformly charged spherical conductors A | vedclass

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

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$2:1$

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(B) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_A = V_B$.Since the potential of a spherical conductor is $V = \frac{KQ}{R}$,we have $\frac{KQ_A}{R_A} = \frac{KQ_B}{R_B}$.This implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{5 \ mm}{10 \ mm} = \frac{1}{2}$.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 $\frac{E_A}{E_B} = \frac{KQ_A / R_A^2}{KQ_B / R_B^2} = \frac{Q_A}{Q_B} 	imes \left(\frac{R_B}{R_A}ight)^2$.Substituting the ratio of charges: $\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}ight) 	imes \left(\frac{R_B}{R_A}ight)^2 = \frac{R_B}{R_A}$.Given $R_A = 5 \ mm$ and $R_B = 10 \ mm$,we get $\frac{E_A}{E_B} = \frac{10}{5} = \frac{2}{1}$.

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