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 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}}\] \[r\], 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 \[{{q}_{A}}={{q}_{C}}=\frac{{{q}_{A}}+{{q}_{B}}}{2}\] \[=\frac{(q/2)+q}{2}\] Hence, electric force acting between \[B\] and \[C\] is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{r}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q/2)(3q/4)}{{{r}^{2}}}\] \[=\frac{3}{8}\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}} \right]=\frac{3F}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
