A) \[\frac{F}{4}\]
B) \[\frac{3F}{4}\]
C) \[\frac{F}{8}\]
D) \[\frac{3F}{8}\]
Correct Answer: D
Solution :
| [d] 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 is being the distance between the charges. |
| When third uncharged conductor A is brought in contact with S, 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 |
| \[{{q}_{A}}={{q}_{C}}=\frac{{{q}_{A}}+{{q}_{B}}}{2}=\frac{(q/2)+q}{2}=\frac{3q}{4}\] |
| Hence, electric force acting between B and C is |
| \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{r}^{2}}}\] |
| \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q/2)(3q/4)}{{{r}^{2}}}\] |
| \[=\frac{3}{8}\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}} \right]=\frac{3F}{8}\] |
You need to login to perform this action.
You will be redirected in
3 sec
