Two identical metallic spheres A and B when p | vedclass

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(B) Let the initial charges on spheres $A$ and $B$ be $q_A = q_B = q$. The initial force is $F = \frac{Kq^2}{r^2}$.When sphere $C$ (initially uncharge

Vedclass · Accepted Answer

$3F / 4$

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(B) Let the initial charges on spheres $A$ and $B$ be $q_A = q_B = q$. The initial force is $F = \frac{Kq^2}{r^2}$.When sphere $C$ (initially uncharged) is placed in contact with $A$,the charge redistributes equally: $q_A' = q_C' = \frac{q}{2}$.Now,sphere $C$ is placed in contact with $B$. The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. Upon separation,each gets $q_B' = q_C'' = \frac{3q/2}{2} = \frac{3q}{4}$.Now,$A$ has charge $\frac{q}{2}$ and $B$ has charge $\frac{3q}{4}$. Sphere $C$ is placed at the midpoint (distance $r/2$ from both).The force on $C$ due to $A$ is $F_1 = \frac{K(q/2)(3q/4)}{(r/2)^2} = \frac{3Kq^2/8}{r^2/4} = \frac{3Kq^2}{2r^2} = \frac{3F}{2}$ (towards $A$).The force on $C$ due to $B$ is $F_2 = \frac{K(3q/4)(3q/4)}{(r/2)^2} = \frac{9Kq^2/16}{r^2/4} = \frac{9Kq^2}{4r^2} = \frac{9F}{4}$ (towards $A$).The net force on $C$ is $F_{net} = F_2 - F_1 = \frac{9F}{4} - \frac{3F}{2} = \frac{9F - 6F}{4} = \frac{3F}{4}$.

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