Repulsion between charged spheres

Question

Two identical conducting spheres carry charges of + 5.0 micro coulombs [5.0 x 10^-6] and -1.0 micro coulombs [ 1.0 x 10^-6] respectively.

The centres of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact The spheres are then returned to their original separation.
What is the ratio of the magnitude of the force on either charge AFTER the spheres are touched to that BEFORE they were touched? Possible answers: 1/2 , 4/5 , 9/5 , 5/1 , 4/9,...... In the equation kQ₁Q₂/L² you would use the initial values of Q₁ and Q₂, i.e. +5 and -1, and r = L to obtain an expression for the initial (attractive) force. The charges become equal after contact, namely +2, and in the new calculation you use these for Q₁ and Q₂, again with r = L. When you take the ratio, the L factors cancel out.

Answer

We take any distance of separation L>>r.

Force between the two charges Q₁ and Q₂ = kQ₁Q₂/L² where k is a constant

The force before contact at distance L is proportional to the product of the two charges on the spheres i.e. +5 x –1 = -5.
The force after contact at distance L is also proportional to the product of the two charges on the spheres i.e. +2 x +2 = +4

Therefore the ratio of the forces after to that before = +4/-5 = -4/5.

Note the negative sign which does not appear in your list of possible answers.