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Calculating the magnitude of the net force bewteen charges

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Nine different charged balls, which we treat as point charges, are arranged in a highly symmetric pattern around a square. Note that the value of Q is 4.00 x 10-6C, and L = 30.0 cm

    Image: http://www.webassign.net/userimages/81224?db=v4net

    2. Relevant equations

    1) What is the magnitude of the net force experienced by the ball with the -3Q charge due to the +3Q and +Q charged balls only?

    2) What is the magnitude of the net force experienced by the ball with the -3Q charge due to the four charged balls with charges of -5Q?

    3 )What is the magnitude of the net force experienced by the ball with the -3Q charge due to the two charged balls with charges of +2Q?

    4) Use your previous three results to find the magnitude of the net force experienced by the ball with the -3Q charge due to the other 8 charged balls.

    3. The attempt at a solution

    1) I got this part correct: I understand how to solve for the force when the point charges are in the straight line

    My work is as follows:

    F3 = F31 + F32
    = k(-3Q)(Q) / L^2 + k(3Q)(+3Q) / 4L^2
    = 5.25k(Q)^2 / L^2
    = 8.39

    2) and 3) :

    Here is where I do not understand how to solve for the force when there are diagonals involved.

    If I am correct in my thinking for 2) then does the -3Q ball experience 2 diagonal forces from the 2 left -5Q balls and a straight force to the right from the 2 right -5Q balls?

    In my calculation I did F53 = 15kq^2 / 3L^2, simplifying to 5kq^2 / L^2 which gives me 7.99 I believe. Since there are essentially 2 diagonal forces, do I multiply this by 2 and then add the other force which is directed right by doing 7.99cos(30 degrees)?

    Thanks for all the input!
     
  2. jcsd
  3. Jan 23, 2009 #2
    The first part of your reasoning is right.

    But consider only the right most column. If we consider this column by itself, why do you think there would be a net force?
     
    Last edited: Jan 24, 2009
  4. Jan 23, 2009 #3
    well i tried putting in 0 for the answer and it did not work because I initially thought the opposing diagonals cancel each other out, but apparently that is not the case.
     
  5. Jan 24, 2009 #4
    Sorry I meant right, why do you think there would be a force on the right most column?

    The two 5q balls on the left do give you a net force that is horizontal -- which can be easily seen if you componentize the forces and use superposition.

    Sorry for the mix up.
     
  6. Jan 24, 2009 #5
    I see what you mean..im not sure how to use superposition in this case since both the forces from the two 5q balls on the left are in opposing diagonals. So the resultant force is directed to the right horizontally correct? I tried finding the F53x and F53y components but i think that both diagonals produced y components that cancel each other out. I am not sure how to proceed with the question.
     
    Last edited: Jan 24, 2009
  7. Jan 24, 2009 #6
    The principal of superposition basically means that the total force is the sum of the forces. You are correct and finding that the y-component of the two balls will cancel out since they are equal and opposite, however you're still left with the x-components.
     
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