Need some clarification on conducting spheres

Question:
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.141 N when their center-to-center separation is 67.0 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0528 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge in coulombs on one of them and (b) the positive charge in coulombs on the other?

I got the right answer, however I am confused on one of the first steps.
I say q1 will be positive and q2 will be negative and Columbs law is set like this:
Code:
F=-(k*q1*q2)/r[SUP]2[/SUP]
My question is why is this negative and for qfinal it is positive when setting up Columbs law.
I hope I was clear enough in this question.

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If I have understood your question correctly, you are wondering why the force is 'negative' when the two spheres are attracting, and 'positive' when they are repelling each other.

What you need to remember is that the force is a vector, and as a result, it has a direction. You have essentially calculated the force experienced by the 'right hand' sphere. Because it is being attracted to the 'left hand' sphere, it feels a force in the '-x' direction, hence the minus sign. There is another term in this formula called the 'unit vector', which dictates which direction the force points, but you don't really need to worry about it; without it, you can neglect the signs of the charges, since you are only interested in finding the magnitude of the force.

If I have understood your question correctly, you are wondering why the force is 'negative' when the two spheres are attracting, and 'positive' when they are repelling each other.

What you need to remember is that the force is a vector, and as a result, it has a direction. You have essentially calculated the force experienced by the 'right hand' sphere. Because it is being attracted to the 'left hand' sphere, it feels a force in the '-x' direction, hence the minus sign. There is another term in this formula called the 'unit vector', which dictates which direction the force points, but you don't really need to worry about it; without it, you can neglect the signs of the charges, since you are only interested in finding the magnitude of the force.
Ok I see now, thanks for the reply and sorry if I wasnt clear.