# Coulombs Law Question

## Homework Statement

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.

## Homework 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?

## 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

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?

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?

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

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

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

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

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

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