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Two fixed charges, find (x,y) coords of third particle.

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  1. Sep 7, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure particle 1 of charge q1 (at origin) and particle 2 of charge q2 = 3q1, are held at separation L on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a)x and (b)y coordinates of particle 3?

    2. Relevant equations
    E = k q1q2/r2

    3. The attempt at a solution
    We know that the third particle will be held at equilibrium between the two particles (so y = 0), so the forces on it from q1 and q2 will cancel out. From this idea we get:

    E13 - E23 = 0
    E13 = E23

    We plug in our relevant equation:

    k q1q3/r2 = k q2q3/(L-r)2

    Plug in and cancel like terms:

    k q1q3/r2 = k q2q3/(L-r)2
    k q1q3/r2 = k 3q1q3/(L-r)2
    k q1q3/r2 = k 3q1q3/(L-r)2
    q1/r2 = 3q1/(L-r)2
    1/r2 = 2q1/(L-r)2

    Attempt to isolate for r:

    (L-r)/r = sqrt(2q1)


    And this is where I get stuck.
     
    Last edited: Sep 7, 2015
  2. jcsd
  3. Sep 7, 2015 #2

    Orodruin

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    You did not do this step correctly.
     
  4. Sep 7, 2015 #3
    I believe I just accidentally left out the division sign. Let me write it out long way just to make sure:

    k q1q3/r2 = k q2q3/(L-r)2
    k q1q3/r2 = k 3q1q3/(L-r)2
    k q1q3/r2 = k 3q1q3/(L-r)2
    q1/r2 = 3q1/(L-r)2
    1/r2 = 2q1/(L-r)2
     
  5. Sep 7, 2015 #4

    Orodruin

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    The last step is still not correct so you need to repeat arithmetic rules. The different sides of the last equation are even dimensionally inconsistent. The left side has dimension length^-2 while the right has dimension charge length^-2.
     
  6. Sep 7, 2015 #5
    Oh wow, I see what you're talking about. I blame my long summer break for this silly arithmetic error. Ok, so I fixed my work and have arrived at:

    q1/r2 = 3q1/(L-r)2
    1/r2 = 3/(L-r)2
    sqrt(1/3) = sqrt(r2/(L-r)2)
    sqrt(1/3) = r/(L-r)
    L(sqrt(1/3) + 1)= r

    So after going through this I just realized my arithmetic is off again... let me figure this out.
     
    Last edited: Sep 7, 2015
  7. Sep 7, 2015 #6
    EDIT: My work was wrong and this post was pointless. Currently working it out again.

    EDIT2: I finally got the answer. I'll post my work for future people that have this same question.

    sqrt(3)/3 = r/(L-r)
    L sqrt(3)/3 - r sqrt(3)/3 = r
    L sqrt(3)/3 = r (1+ sqrt(3)/3)
    r = (L sqrt(3)/3)/(1+ sqrt(3)/3)
     
    Last edited: Sep 7, 2015
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