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(B) Initially, the force between spheres $P$ and $S$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2} = 16 \,N$.When sphere $R$ (uncharged) is brough

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$6$

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(B) Initially, the force between spheres $P$ and $S$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2} = 16 \,N$.When sphere $R$ (uncharged) is brought in contact with sphere $P$ (charge $Q$), the charge is shared equally between them because they are identical. Thus, the new charge on $P$ is $Q_P' = \frac{Q+0}{2} = \frac{Q}{2}$. The charge on $R$ becomes $\frac{Q}{2}$.Next, sphere $R$ (now with charge $\frac{Q}{2}$) is brought in contact with sphere $S$ (charge $Q$). The total charge is shared equally between them. Thus, the new charge on $S$ is $Q_S' = \frac{Q + Q/2}{2} = \frac{3Q/2}{2} = \frac{3Q}{4}$.The new force of repulsion between $P$ and $S$ is $F' = k \frac{Q_P' Q_S'}{r^2} = k \frac{(Q/2)(3Q/4)}{r^2} = \frac{3}{8} \left( k \frac{Q^2}{r^2} ight)$.Substituting the initial force $F = 16 \,N$, we get $F' = \frac{3}{8} 	imes 16 \,N = 6 \,N$.

Two identical conducting spheres $P$ and $S$ with charge $Q$ on each, repel each other with a force $16 \,N$. $A$ third identical uncharged conducting sphere $R$ is successively brought in contact with the two spheres. The new force of repulsion between $P$ and $S$ is: (in $\,N$)

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