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Finding third charge coordinates in an equilibrium position

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Here are the problem statement and the solution. I'm stuck at where the book suggests the formulas for the x and y coordinations (highlighted in yellow) of the third charge. Any explanations or proof on how they came to the conclusion for the third charge coordinations would be much appreciated. Is there an easier solution for this problem?

    fKbmPI0.png

    The given solution:

    GtmJTjO.png

    2. Relevant equations
    x3=x2-rcosθ
    y3=y2-rsinθ
    3. The attempt at a solution
    I think it must have something to do with the "point-slope" form of the equation of a straight line, but stuck in finding a reasonable proof.
     
    Last edited by a moderator: Feb 17, 2017
  2. jcsd
  3. Feb 17, 2017 #2

    BvU

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    Lazy, lazy, eh ? Perhaps you can at least make a drawing of the situation and post it, together with a non-empty attempt at solution. You know the guidelines at PF I hope !
     
  4. Feb 17, 2017 #3

    haruspex

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    You highlighted collinear, so I assume you do not understand why they must be. Draw yourself a diagram of three charges not collinear and consider the forces on one of them. Can those forces be in balance?
     
  5. Feb 18, 2017 #4
    OK, I've progressed a bit. Now, I don't know why there is also a negative sign before rsinθ in y3=y2-rsinθ.
    KcXU5G2.png
     
  6. Feb 18, 2017 #5

    haruspex

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    What does the book solution state in parentheses immediately after that? Does that accord with the way you have drawn θ?
     
  7. Feb 18, 2017 #6
    So it means that the negative sign is because of the fact that sin(-θ)=-sin(θ)?
     
  8. Feb 18, 2017 #7

    BvU

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    Theta is the angle wrt the negative x-direction. That's all.
     
  9. Feb 18, 2017 #8

    haruspex

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    Yes.
    They define θ as the angle the vector from q3 towards q1 makes to +ve x axis. By convention, that would mean anticlockwise from that axis. Your diagram shows θ measured clockwise from the axis, so your θ is minus their θ.
     
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