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Three Charges in a Triangle

  1. Jan 18, 2015 #1
    1. The problem statement, all variables and given/known data

    Three 3.0 g balls are tied to 80-cm-long threads and hung from a single fixed point. Each of the balls is given the same charge q. At equilibrium, the three balls form an equilateral triangle in a horizontal plane with 20 cm sides.

    2. Relevant equations

    F=kq1q2/r^2

    F (of 1 on 2) = F (of 2 on 1)

    3. The attempt at a solution

    Since they all have the same charge, I figured that the forces would be equal at each point. All points are in the same plane, so I assumed they only have x and y components. I tried using F=kq1q2/r^2=kq2q3/r^2=kq1q3/r^2 but it ended up a huge mess with all the charges canceling. But they all have the same charge? So should I use F=k2q/r^2 and do something with that?
     
  2. jcsd
  3. Jan 18, 2015 #2

    mfb

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    The forces are equal in magnitude, but not in direction.
    Due to the symmetry, it is sufficient to consider the position of a single charge. What are all forces acting on it, including their directions?

    What is "k2q"?
     
  4. Jan 18, 2015 #3
    Well on each charge, since they are positioned in an equilateral triangle, they all have 2 forces acting on them, both at 30 degrees (I think?) and one would be at +30 degrees and one would be at -30 degrees.

    It was supposed to be kq^2, from the F=(k*q1*q2)/r^2 equation, but since q1=q2 we could just use q^2.
     
  5. Jan 18, 2015 #4

    mfb

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    Okay.
    Right, but that is just the magnitude of the charge.
     
  6. Jan 18, 2015 #5
    Sure, so I have the magnitude. And each charge is sitting in a triangle with sides 20cm. So, if we put the first point at say, what I'll make my origin, then I have another point 20 cm to the right, so now its coordinates are (20,0) and the third charge would be at about... (10, 17.3)? Okay. So how do their relative positions help?
     
  7. Jan 18, 2015 #6

    mfb

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    You listed the two from the electrostatic repulsion, but there are two more. Those objects are not floating in free space.
     
  8. Jan 18, 2015 #7
    OH! Gravity and the tension in the strings!
     
  9. Jan 19, 2015 #8

    mfb

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    Right. And if you add all together, ...
     
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