Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart . The spheres are brought into contact and then taken to original positions . Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force . The ratio between initial charges on the spheres is

Options

A. a)`-(3+sqrt(8))` only
B. b)`3+sqrt(8)` only
C. c)`-(3+sqrt(8)or(-3+sqrt(8))`
D. d)`+sqrt(3)`

Answer

C

Explanation

Since the two charges (spheres)attract , they will be opposite in sign , ie ,`q_(1)and-q_(2)`.
Force , `F=(1)/(4piepsilon_(0))(q_(1)xx-q_(2))/(d^(2))`
After touching , charge on each will be `(q_(1)-q_(2))/(2)`.
New force , `F.(1)/(4piepsilon_(0))((q_(1)-q_(2))/(2))^(2)xx(1)/(d^(2))`
Given `|F|=|F.|`
On solving by quadratic eqadratic equations , we get
`(q_(1))/(q_(@))=-(3+sqrt(8))or(-3+sqrt(8))`