Coulombs Law Question

  • Thread starter paperdoll
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  • #1
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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
 

Answers and Replies

  • #2
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
69
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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 :confused:
 
  • #5
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 :confused:

Yes, the charges would evenly distribute themselves over the connection made by touching.
 
  • #6
69
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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
 
  • #7
9
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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
 

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