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Point charges on equilateral triangle no idea what's wrong.

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data

    These are really frustrating because they seem so simple, I work them all out without a hitch only to find that I'm wrong half the time and left wondering why.

    In fig 21-26a, particles 1 and 2 have charge 20 microcoulombs each and are held at separation distance d = 1.5 meters.

    (1 is located above 2)

    a.) What is the magnitude of the electrostatic force on 1 due to 2?
    b.) Particle 3 is also of charge 20 microcoulombs and is placed to the right-center of 1 and 2, forming an equilateral triangle with 1 and 2. Find the net electrostatic force on 1 due to 2 and 3.

    2. Relevant equations



    3. The attempt at a solution

    The first part, I have correct.

    [itex]F = kq_{1}q_{2}r^{-2} = (8.99x10^{9})(20x10^{6})(20x10^{6})(1.5)^{2} = 1.6 N[/itex]

    The second part, I do not have correct.

    Since the magnitude of the force from 2 is 1.6 N and it is located below 1, the vector of the force is just 1.6j.

    Since the particles make an equilateral triangle, every angle is 60*. Since 3 is to the right, the positive angle to 1 from 3 is 120*.

    Therefore the unit vector that points from 3 to 1 is cos(120)i + sin(120)j.

    Since the magnitude of force from 3 to 1 is also 1.6, the vector of the force due to 3 is:

    [itex]1.6cos(120)i + 1.6sin(120)j = -.8i + 1.38j[/itex]

    Adding the j components, the vector of all force on 1 is:

    -.8i + 2.98j which has a magnitude of roughly 3.09 newtons.



    The book answer is 2.77. I don't even understand how their result makes sense. 2 and 3 are purely additive, they are both below 1 and 2 has no x component while 3 does.

    Help? TIA
     
  2. jcsd
  3. Jul 21, 2012 #2

    TSny

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    Hi, 1MileCrash. 120o??
     
  4. Jul 21, 2012 #3
    Was my explanation poor?
     
  5. Jul 21, 2012 #4

    TSny

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    I just wanted you to focus on the 120o. I don't think 120o is the angle you want for constructing the unit vector from 3 to 1. I'm sure you've drawn a sketch, but you might look at it again.
     
  6. Jul 21, 2012 #5
    Nevermind, I get you!

    It's not 120* because I was subtracting the full 60 from 180 when I should only be subtracting half of that 60 from 180, to get 150. Right?
     
  7. Jul 21, 2012 #6

    TSny

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    Try it and see :wink:
     
  8. Jul 21, 2012 #7
    Fair enough, thank you sir!
     
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