Two equally charged metal spheres A and B rep | vedclass

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Question

(A) Let the initial charge on both spheres $A$ and $B$ be $q$. The distance between them is $r$. The initial force is $F = k \frac{q^2}{r^2} = 4 	ime

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the initial charge on both spheres $A$ and $B$ be $q$. The distance between them is $r$. The initial force is $F = k \frac{q^2}{r^2} = 4 	imes 10^{-5} \ N$. When uncharged sphere $C$ touches $A$,the charge on $A$ becomes $q_A = q/2$ and the charge on $C$ becomes $q_C = q/2$. Now,sphere $C$ is placed at the midpoint between $A$ and $B$. The distance of $C$ from both $A$ and $B$ is $r/2$. The force exerted by $A$ on $C$ is $F_{AC} = k \frac{(q/2)(q/2)}{(r/2)^2} = k \frac{q^2/4}{r^2/4} = k \frac{q^2}{r^2} = 4 	imes 10^{-5} \ N$ (directed towards $B$). The force exerted by $B$ on $C$ is $F_{BC} = k \frac{(q)(q/2)}{(r/2)^2} = k \frac{q^2/2}{r^2/4} = 2k \frac{q^2}{r^2} = 2(4 	imes 10^{-5}) = 8 	imes 10^{-5} \ N$ (directed towards $A$). The net force on $C$ is $F_{net} = F_{BC} - F_{AC} = 8 	imes 10^{-5} - 4 	imes 10^{-5} = 4 	imes 10^{-5} \ N$. Since $F_{BC} > F_{AC}$,the net force is directed 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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