Two identical conducting sphere 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 their original positions. Now, the two sphere repel each other with a force whole magnitude is equal to the initial attractive force. The ratio between initial charges on the spheres is

Options

A. 2
B. 3
C. `3//2`
D. Information insufficient

Answer

B

Explanation

Metal spheres attract initially hence, they must be carrying charges of opposite nature. Let these charges be `Q_1` and `-Q_2`, Using Coulomb.s law we can write the following:
`F_1=1/(4piepsilon_0)(Q_1Q_2)/(0.5)^2`=0.108 …(i)
When spheres are brought in contact then amount of charge on both becomes `(Q_1- Q_2)//2`. When these are again placed at the same distance then the force between them can be written as follows:
`F_2=1/(4piepsilon_0)(((Q_1-Q_2)/2)^2)/(0.5)^2` =0.036 …(ii)
Dividing equation (i) and (ii) we get the following :
`(Q_1Q_2)/(Q_1-Q_2)^2 =0.108/(4xx0.036)=3/4`
`4Q_1Q_2=3Q_1^2+3Q_2^2-6Q_1Q_2`
`10Q_1Q_2=3Q_1^2 +3Q_2^2`
`10=3Q_1/Q_2+3Q_2/Q_1`
`10=3x+3/x`, where `x=Q_1/Q_2`
`10x=3x^2+3 rArr 3x^2 -10x +3 =0`
`rArr 3x^2-9x-x+3=0`
`rArr 3x(x-3)-(x-3)=0`
`rArr` (x-3)(3x-1)=0
`rArr` x=3 and 1/3
Here both the answers are basically the same that one of the charges must be having three times the magnitude of other. It can be 3Q and -Q or you can say Q and -3Q.