Two isolated conducting spheres S_{1} and S_{ | vedclass

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Question

(A) When two conducting spheres are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_{1}'$ and $Q_{2}

Vedclass · Accepted Answer

$6\, \mu C$ and $3\, \mu C$

Vedclass · Answer

(A) When two conducting spheres are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_{1}'$ and $Q_{2}'$. According to the law of conservation of charge,the total charge remains constant: $Q_{1}' + Q_{2}' = Q_{1} + Q_{2} = 12\, \mu C - 3\, \mu C = 9\, \mu C$. Since the potentials are equal $(V_{1} = V_{2})$,we have: $\frac{K Q_{1}'}{R_{1}} = \frac{K Q_{2}'}{R_{2}} \Rightarrow \frac{Q_{1}'}{2R/3} = \frac{Q_{2}'}{R/3}$. This simplifies to $Q_{1}' = 2 Q_{2}'$. Substituting this into the conservation equation: $2 Q_{2}' + Q_{2}' = 9\, \mu C \Rightarrow 3 Q_{2}' = 9\, \mu C \Rightarrow Q_{2}' = 3\, \mu C$. Then,$Q_{1}' = 2 	imes 3\, \mu C = 6\, \mu C$. Thus,the final charges are $6\, \mu C$ and $3\, \mu C$.

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