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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

Answer»

`SQRT(15/16)`
`sqrt(16/15)`
`sqrt(20/12)`
`sqrt(12/20)`

Solution :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)`


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