Two identical conducting spheres,having charg | vedclass

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(B) Let the initial charges be $Q_1$ and $Q_2$. Since they have opposite signs,let $Q_1 = q$ and $Q_2 = -k q$ (where $k > 0$).Initial force $F_1 = \fr

Vedclass · Accepted Answer

$\pm 1.0 	imes 10^{-6} \ C$ and $\mp 3.0 	imes 10^{-6} \ C$

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(B) Let the initial charges be $Q_1$ and $Q_2$. Since they have opposite signs,let $Q_1 = q$ and $Q_2 = -k q$ (where $k > 0$).Initial force $F_1 = \frac{k_e |Q_1 Q_2|}{r^2} = 0.108 \ N$.$0.108 = \frac{9 	imes 10^9 |Q_1 Q_2|}{(0.5)^2} \implies |Q_1 Q_2| = \frac{0.108 	imes 0.25}{9 	imes 10^9} = 3 	imes 10^{-12} \ C^2$.Since they have opposite signs,$Q_1 Q_2 = -3 	imes 10^{-12} \ C^2$.When connected by a wire,the total charge $Q_{total} = Q_1 + Q_2$ is shared equally between the identical spheres.Each sphere now has charge $Q' = \frac{Q_1 + Q_2}{2}$.Final force $F_2 = \frac{k_e (Q')^2}{r^2} = 0.036 \ N$.$0.036 = \frac{9 	imes 10^9}{0.25} \left( \frac{Q_1 + Q_2}{2} ight)^2$.$\left( \frac{Q_1 + Q_2}{2} ight)^2 = \frac{0.036 	imes 0.25}{9 	imes 10^9} = 1 	imes 10^{-12} \ C^2$.$Q_1 + Q_2 = \pm 2 	imes 10^{-6} \ C$.Solving $Q_1 + Q_2 = 2 	imes 10^{-6}$ and $Q_1 Q_2 = -3 	imes 10^{-12}$,we get $Q_1 = 3 	imes 10^{-6} \ C$ and $Q_2 = -1 	imes 10^{-6} \ C$ (or vice versa).

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