What Is the Ratio of Electrostatic Forces After Introducing a Third Sphere?

[Geekster]

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters. The electrostatic force acting on sphere 2 due to sphere 1 is F and the force acting on sphere 1 due to sphere 2 is -F'. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1, then to sphere 2, and finally removed. What is the ratio F'/F?

So when the thrid sphere touches the first sphere it takes with it half the total charge of sphere 1. Then touching it to sphere two it takes only 1/4 of the charge.

So now sphere 1 has 1/2 the original charge and sphere 2 has 3/4 of the original charge. According to Coulomb's law the force of sphere 1 on sphere 2 is going to be $$k|q_1q_2|/r^2$$.

So wouldn't the force of sphere 1 on sphere 2 be the same as the force of sphere 2 on sphere 1? And if that is the case, then the ratio would be -1, right?

 

[Geekster]

So did I do everything right? Or am I so bad that no one wants to respond?