Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting spheres, C,is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to :

Options

A. `(3F)/4`
B. `F/2`
C. `(3F)/8`
D. F

Answer

C

Explanation

Force between A and B spheres
`F=(kq^2)/r^2`
When A and C are touched then charged on both will be equally distributed and that is `q/2`
`q_A=q/2, q_C=q/2`
Similarly, when B and C are touched
`q_B=(q+q/2)/2 =(3q)/4`
`F.=(kq_Aq_B)/r^2=(kxxq/2xx(3q)/4)/r^2 = 3/8 (kq^2)/r^2 =3/8F`