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Homework Help: Three Point Charges, Find x-component

  1. Jan 16, 2015 #1


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    1. The problem statement, all variables and given/known data
    In the figure, three point charges are arranged in a equilateral triangle of side a= 1.40×10-2 m. Sketch the field lines due to + Q and -Q, where Q = 2.50×10-5 C , and from them determine the direction of the force that acts on +q = 5.00×10-8 C because of the presence of the other two charges. What is the x-component of the force? What is the y-component of the force?

    2. Relevant equations

    3. The attempt at a solution
    I know that the y-component is 0 N.
    Someone posted this problem in yahoo answers, and I tried to follow it. Something went wrong along the way, but here is the work from that solution:

    *+Q=1, -Q=2, +q=3*

    F1,3=(8.99E9)(5.00E-8)(2.50E-5)/(1.40E-2)=0.803 N
    F2,3=(8.99E9)(5.00E-8)(-2.50E-5)/(1.40E-2)=-0.803 N

    F1,3x=0.803(cos(30 degrees))=0.6951 (right)
    F1,3y=0.803(cos(60))=0.4013 (down)

    F2,3x=-0.803(cos(30))=-0.6951 (left)
    F2,3Y=-0.803(cos(60))=-0.4013 (down

    Here, I have the problem where if you combine x and y values, the answer is 0 for both of them.
    Where did I go wrong, or is all of this wrong?
  2. jcsd
  3. Jan 16, 2015 #2

    Doc Al

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    Staff: Mentor

    Get rid of that negative sign. Use Coulomb's law to find the magnitude of the force, then use the signs to find the direction. Draw yourself a diagram of the forces on q.
  4. Jan 20, 2015 #3


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    Sorry for not replying in a while!
    In the work I did, I completely forgot to square a, making all my numbers wrong.

    ((9E9)(5.00E-8)(2.50E-5))/(1.40E-2)^2=57.33 N (this is the x-component)

    Thank you!
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