A) \[\frac{F}{4}\]
B) \[\frac{3F}{4}\]
C) \[\frac{F}{8}\]
D) \[\frac{3F}{8}\]
Correct Answer: D
Solution :
Let the spherical conductors B and C have same charge as q. The electric force between them is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}}\] where r, being the distance between them. When third uncharged conductor A is brought in contact with B, then charge on each conductor \[{{q}_{A}}={{q}_{B}}=\frac{{{q}_{A}}+{{q}_{B}}}{2}\] \[=\frac{0+q}{2}=\frac{q}{2}\] When this conductor A is now brought in contact with C, then charge on each conductor \[{{q}_{A}}={{q}_{C}}=\frac{{{q}_{A}}+{{q}_{C}}}{2}\] \[=\frac{(q/2)+q}{2}\] \[=\frac{3q}{4}\] Hence, electric force acting between B and C is \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{r}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q/2)(3q/4)}{{{r}^{2}}}\] \[=\frac{3}{8}\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}} \right]=\frac{3F}{8}\]
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