Two equally charged,identical metal spheres A | vedclass

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(A) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = k\frac{Q^2}{r^2}$.When sphere $C$ (uncharged) is brought in contac

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$F$

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(A) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = k\frac{Q^2}{r^2}$.When sphere $C$ (uncharged) is brought in contact with sphere $A$ (charge $Q$),the charge redistributes equally between them. Thus,the new charge on $A$ is $Q_A = Q/2$ and on $C$ is $Q_C = Q/2$. Sphere $B$ remains with charge $Q_B = Q$.Sphere $C$ is placed at the midpoint,so the distance from $A$ is $r/2$ and from $B$ is $r/2$.The force on $C$ due to $A$ is $F_A = k\frac{(Q/2)(Q/2)}{(r/2)^2} = k\frac{Q^2/4}{r^2/4} = k\frac{Q^2}{r^2} = F$. This force acts away from $A$ (repulsive).The force on $C$ due to $B$ is $F_B = k\frac{(Q)(Q/2)}{(r/2)^2} = k\frac{Q^2/2}{r^2/4} = 2k\frac{Q^2}{r^2} = 2F$. This force acts away from $B$ (repulsive).Since $F_A$ and $F_B$ act in opposite directions,the net force on $C$ is $F_{net} = |F_B - F_A| = |2F - F| = F$.

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