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(D) When two conducting spheres are connected,charge flows until they reach the same electric potential $V$.Let the final charges on the spheres of ra

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$\frac{4q}{3}$

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(D) When two conducting spheres are connected,charge flows until they reach the same electric potential $V$.Let the final charges on the spheres of radius $R$ and $2R$ be $q_1$ and $q_2$ respectively.By conservation of charge,$q_1 + q_2 = q + (-2q) = -q$.The potential of the spheres is given by $V = \frac{k q_1}{R} = \frac{k q_2}{2R}$,which implies $q_2 = 2q_1$.Substituting this into the conservation equation: $q_1 + 2q_1 = -q$,so $3q_1 = -q$,which gives $q_1 = -\frac{q}{3}$.The charge that flows from the first sphere is the change in its charge: $\Delta q = q_{initial} - q_{final} = q - (-\frac{q}{3}) = \frac{4q}{3}$.

$A$ conducting sphere of radius $R$ carrying a charge $q$ is joined to a conducting sphere of radius $2R$ carrying a charge $-2q$. The charge flowing between them will be

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