Two equally charged metal spheres A and B rep | vedclass

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(A) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{d^2} = 4 	imes 10^{-5} ~N$. When sphere $C$ (uncharg

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$4 	imes 10^{-5} ~N$ from $C$ to $A$

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(A) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{d^2} = 4 	imes 10^{-5} ~N$. When sphere $C$ (uncharged) is touched to sphere $A$,the charge redistributes equally between them. Thus,the new charge on $A$ is $q_A' = q/2$ and the charge on $C$ is $q_C = q/2$. The charge on $B$ remains $q_B = q$. Sphere $C$ is placed at the midpoint,so its distance from both $A$ and $B$ is $d/2$. The force on $C$ due to $A$ is $F_A = \frac{k (q/2)(q/2)}{(d/2)^2} = \frac{k q^2}{d^2} = F = 4 	imes 10^{-5} ~N$ (directed away from $A$,i.e.,towards $B$). The force on $C$ due to $B$ is $F_B = \frac{k (q)(q/2)}{(d/2)^2} = 2 \frac{k q^2}{d^2} = 2F = 8 	imes 10^{-5} ~N$ (directed away from $B$,i.e.,towards $A$). The net force on $C$ is $F_{net} = F_B - F_A = 8 	imes 10^{-5} - 4 	imes 10^{-5} = 4 	imes 10^{-5} ~N$ directed towards $A$ (from $C$ to $A$).

Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 \times 10^{-5} ~N$. Another identical uncharged sphere $C$ is touched to $A$ and then placed at the midpoint of the line joining the spheres $A$ and $B$. The net electric force on the sphere $C$ is

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