Two identical very small metal spheres, A and B, that are electrically neutral are fixed at a distance R On insulating handles. A third identical sphere C carrying charge Q is also connected to an insulating handle. C ia first brought into contact with A and B and then moved away from these two spheres. It is found that A and B apply `F_1` force on each other.
Now C with the remaining charge is again brought into contact with A and then moved away from A and B. It is found that A and B apply `F_2` force on each other. Now C is further brought into contact with B and then moved away from A and B. It is found that A and B apply `F_3` force on each other Finally B is moved closer to A at a distance r, such that the force between them again becomes `F_1`.
The magnitude of r/R is

Options

A. `sqrt(15/16)`
B. `sqrt(16/15)`
C. `sqrt(20/12)`
D. `sqrt(12/20)`

Answer

A

Explanation

Now the distance between A andB is changed to r so that new force between A and B becomes equal to `F_1`. Hence we can write the following equation for the new distance between the charges.
`15/128 1/(4piepsilon_0) Q^2/r^2 =1/8xx1/(4piepsilon_0) Q^2/R^2 rArr 15/(128r^2) =1/(8R^2) rArr r=R sqrt(15/16)`