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Coulomb's Law (3 charges in equilateral triangle)

  1. Feb 25, 2014 #1
    Three charged particles are placed at the corners of an equilateral triangle of side L = 1.74 m.
    The charges are q1 = 3.63 µC, q2 = −8.05 µC, and q3 = −6.31 µC. Calculate the magnitude and direction (counterclockwise from the positive x axis) of the net force on q1 due to the other two charges.

    rni6vU0.gif


    First, I converted the charges from µC to C

    q1 = 3.63e-6 C
    q2 = -8.05e-6 C
    q3 = -6.31e-6 C

    Then I used Coulomb's law to figure out the forces on q1 from the other 2 charges

    F21 = the force applied to charge 1 by charge 2
    F31 = the force applied to charge 1 by charge 3
    r = the distance between each charge = 1.74m

    F21 = K*q1*q2 / r2
    = (8.998e9)(3.63e-6)(-8.05e-6) / 1.742
    = -0.0868 N

    F31 = (8.998e9)(3.63e-6)(-6.31e-6) / 1.742
    = -0.0681 N

    Now because these forces applied are on the diagonal of that equilateral triangle, I must find the x and y components of the forces. The angle inside any equilateral triangle is 60 degrees, so I'm using 60 for my trig here.

    x

    (-0.0868 * cos60) + (-0.0681 * cos60) = -0.0774 N

    y

    (-0.0868 * sin60) + (-0.0681 * sin60) = -0.134 N

    To find the magnitude, I've been adding the square of the x and y component, then taking the square root of it.

    sqrt[(-0.0774 N)2 + (-0.134 N)2] = 0.155 N

    I'm not sure if this is correct, nor am I sure how to get the direction in degrees. This method of finding the force makes sense to me but I need to submit both the magnitude and direction at the same time to get the answer correct.

    Please help!
     
  2. jcsd
  3. Feb 25, 2014 #2

    ehild

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    The magnitude of the forces is positive, the - sign indicates that the force acts towards the other charge. Draw the individual forces. Do the x components have the same (negative) sign?


    ehild
     
  4. Feb 26, 2014 #3
    Not quite sure I understand what you mean by - sign indicating that the force acts towards the other charge. Care to explain a little further?
     
  5. Feb 26, 2014 #4

    ehild

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    You wrote that F21 = -0.0868 N and F31= -0.0681 N. Are those forces parallel? What does the negative sign mean?
    You also considered both x components negative. With respect to what?

    ehild
     
  6. Feb 26, 2014 #5
    I was just going under what the question gave me in terms of q2 and q3's charges. They are both negative charges, therefor wouldn't the Force come out negative just in terms of signage? I suppose this is where I'm confused about the question.
     
  7. Feb 26, 2014 #6

    SammyS

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    Consult the rules of this Forum regarding how long to wait before "Bumping" your thread.
     
  8. Feb 26, 2014 #7

    SammyS

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    All that the negative sign means is that the force is attractive.

    You need to use the geometry of the problem to get the components of the Force(s) .
     
  9. Feb 26, 2014 #8

    ehild

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    The forces have direction, (see picture) determine the x and y components accordingly.


    ehild
     

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