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. F
D. `(3F)/(8)`

Answer

D

Explanation

Spheres A and B carry equal charge say .q.
`:.` Force between them , ` F = (k q q )/( r^(2))`
When A and C are touched, charge on both ` q_(A) = q_(C) = (q)/(2)`
Then when B
`q_(B) = ((q)/(2)+q)/(2)(3q)/(4)`
Now, the force between charge `q_(A) and q_(B)`
`F. = (kq_(A) q_(B))/(r^(2)) = (k xx (q)/(2)xx (3q)/(4))/(r^(2)) = (3 )/(8) (kq^(2))/( r^(2)) = (3)/(8) F`