Need some clarification on conducting spheres

[RandellK02]

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:

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.

 

[davo789]

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.

 

[RandellK02]

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.

 

[davo789]

Glad I could help! :)

 

[rwisz]

I can provide some clarification on conducting spheres and the use of Coulomb's law in this scenario. First of all, it is important to understand that Coulomb's law describes the electrostatic force between two charged particles (in this case, the conducting spheres). The negative sign in the equation represents the attractive force between opposite charges, while a positive sign represents the repulsive force between like charges.

In this scenario, the spheres are initially attracting each other, which means that one sphere must have a positive charge and the other must have a negative charge. This is why q1 is positive and q2 is negative in the equation. When the spheres are connected by a conducting wire, the charges are able to flow between them, equalizing their charges and resulting in a repulsive force. In this case, both spheres will have the same charge, either positive or negative.

To determine the initial charges on the spheres, we can use the equation for Coulomb's law and solve for the charges. We know the force, the distance between the spheres, and the value of the Coulomb constant (k). By plugging in these values and solving for the charges, we can determine their initial values.

I hope this explanation helps to clarify the use of Coulomb's law in this scenario and why one charge is positive and the other is negative. It is important to remember that the signs in Coulomb's law represent the type of force (attractive or repulsive) between the charged particles, not the actual charge itself.