Coulombs Law Question

1. Oct 16, 2011

paperdoll

1. The problem statement, all variables and given/known data
I don't understand how to apply coulombs law when there are 3 balls, giving each other charge. I know that the answer for part a) is 0.02N and part b) is 0.005 N but have no idea how to get these answers.

2. Relevant equations
Two identical plastic balls A and B with a metallic coating are given identical but opposite charges and placed a fixed distance apart. There is found to be an attractive force of 0.04N between them. A third identical, but initially uncharged ball, C is now brought into contact with ball A and then removed.
a) what will be the force between A and B now
b) if ball c, still with the charge it obtained from ball a is now brought into contact with ball b and then removed, what will the force be between a and b now?

3. The attempt at a solution
For part a) I tried substituting -1 and +1 for the charge but I still don't understand how the ball's A and C have transferred charge.
And for part b) I need to understand how to get part a) first

2. Oct 16, 2011

NewtonianAlch

Think of charge distribution.

When you first touch ball C to ball A, some of the charge from ball A will be distributed through C.

How much do you think that would be proportionally if ball C is initially uncharged?

3. Oct 17, 2011

paperdoll

erm...hmm when you first touch ball A to C would the total charge from A half-out with C? so now both ball is the same charge? I'm not sure if electricity works in the same way as heating and cooling

4. Oct 17, 2011

da_steve

5. Oct 17, 2011

NewtonianAlch

Yes, the charges would evenly distribute themselves over the connection made by touching.

6. Oct 17, 2011

paperdoll

so now that both balls are the same charge, the charge of the ball A is halfed? so thats why the 0.04 N --> 0.02 N for part a) of the question

7. Oct 17, 2011

da_steve

Yeah

F=(1/(4∏ε))*(q1*q2/r2)

so when q1 is halved, F is halved

say A had and original charge of q0

A and C now have a charge of 1/2 q0

C touches B the total charge on both is q0 + 1/2 q0 as equal surface area
=> evenly distributed again, then sub into your formula