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Coulombs Law - 3 point charges

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Three identical point charges A, B, and C are located in the shape of an equilateral triangle with sides of length 15 cm. What is the net force on B if each charge has a magnitude of -5.0 x 10-3 C?



    2. Relevant equations

    coulombs law equation

    3. The attempt at a solution
    (please see the attached scanned page and attached diagram)
    I think that i am doing something wrong the angle...hope you guys can help.
     

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  2. jcsd
  3. May 13, 2009 #2

    LowlyPion

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    Without seeing your diagram, I would hope that you are treating the E-field as vectors. The best way to calculate the Sum is to separate them into the x,y components and then add and then report the resultant.
     
  4. May 13, 2009 #3
    did you see my work shown? Is it not opening up?
    I trying to solve the triangle to find the net force. I think that is where I may be getting the angle wrong. It is not suppose to be 60 degrees.
     
  5. May 13, 2009 #4

    LowlyPion

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    Until images are approved, they are not generally available.

    You of course do have 60° angles.

    You may want to choose your axis as the bisector of the opposite side if you did not already. That way you are likely only interested in the components of the 2 other charges directed along this bisector. That should reduce your calculation to simply be 2*cos30*|F| calculated from one of them alone.
     
  6. May 13, 2009 #5
    okay, so i know that the angle for the original triangle is 60 degrees. And by using coulombs law to find Fbc and Fba we can create another triangle to find the net force. I wish there was a way i could draw it out in here, but i guess I will have to wait for the approval. Basically the net force on B i get is 1.0 x 10^7 N.
     
  7. May 13, 2009 #6

    LowlyPion

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    That looks only like |F| between either of the Fab or Fcb

    When you add them as vectors I think you should get something more like

    2*Cos30*|F| = 2*.866*1*107 = 1.732 * 107

    directed away from the other 2 charges along the line that is the ⊥ bisector of the line AC.
     
  8. May 13, 2009 #7
    But don't you have to add the vectors head to tail? and then using that we can use cosine law to solve or vector components like you said earlier.
     
  9. May 13, 2009 #8

    LowlyPion

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    That's one way to do it.

    I didn't use the law of cosines to solve however.

    I simply exploited the symmetry of the charges noting that since they are equal and equidistant, and on opposite sides of the bisector, that their contributions ⊥ TO the bisector cancel, and along the bisector they will add. Since Cos30 is the side in the direction of the bisector, then it is Cos30*|F| for each or 2*Cos30*|F|.
     
  10. May 13, 2009 #9
    when will the attachments be approved? I think it will help me if you maybe see what I did for my solution.
    Thanks!
     
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