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Question

(B) Let the large sphere have radius $R$ and initial charge $Q$. Let the small spheres have radius $r$.When the first sphere is connected,the potentia

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$\frac{q_2^2}{q_1}$

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(B) Let the large sphere have radius $R$ and initial charge $Q$. Let the small spheres have radius $r$.When the first sphere is connected,the potential of the large sphere and the small sphere must be equal. Let the new charge on the large sphere be $Q'$ and on the small sphere be $q_1$. Since the spheres are far apart,we can ignore the potential due to the other sphere: $\frac{q_1}{r} = \frac{Q'}{R}$.Also,by conservation of charge,$q_1 + Q' = Q$.Substituting $Q' = Q - q_1$,we get $\frac{q_1}{r} = \frac{Q - q_1}{R}$,which simplifies to $q_1 = \frac{Q r}{R + r}$.The remaining charge on the large sphere is $Q' = Q - q_1 = Q - \frac{Q r}{R + r} = \frac{Q R}{R + r}$.When the second sphere is connected,the large sphere has charge $Q'$. Let the new charge on the large sphere be $Q''$ and on the second sphere be $q_2$. Then $\frac{q_2}{r} = \frac{Q''}{R}$ and $q_2 + Q'' = Q'$.Solving this gives $q_2 = \frac{Q' r}{R + r} = \frac{Q R r}{(R + r)^2}$.The remaining charge on the large sphere is $Q'' = Q' - q_2 = Q' - \frac{Q' r}{R + r} = \frac{Q' R}{R + r} = \frac{Q R^2}{(R + r)^2}$.When the third sphere is connected,the charge $q_3$ acquired is $q_3 = \frac{Q'' r}{R + r} = \frac{Q R^2 r}{(R + r)^3}$.Now,consider the ratio $\frac{q_2^2}{q_1} = \frac{[\frac{Q R r}{(R + r)^2}]^2}{\frac{Q r}{R + r}} = \frac{Q^2 R^2 r^2}{(R + r)^4} \cdot \frac{R + r}{Q r} = \frac{Q R r}{(R + r)^3} = q_3$.

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