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ElectroStatic Question

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Say there is a charge of +3 at the origin and a charge of -7 at .5m Where would a third charge of arbitrary sign have to be for equilibrium to be reached?


    2. Relevant equations



    3. The attempt at a solution
    so I've widdled this down to 3 / r^2 = -7/(.5+r)^2 but am having problems solving for r. Algebra is tough!!! anyone want to give me tips here? In the equation I have I already divided out the third charge and the k term.
     
    Last edited: Jan 30, 2014
  2. jcsd
  3. Jan 30, 2014 #2

    haruspex

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    It's particularly tough when you try to make a determinedly positive term equal an insistently negative one :wink:. The signs depend on whether the test charge is placed left or right of the given charges.
     
  4. Jan 30, 2014 #3
    Ahh right! So I could just do 3/r^2 = 7/(.5+r)^2?
    Following up would give...
    sqrt(3/7) = (.5+r)/r??
    sqrt(3/7)*r = .5 + r
    0 = .5 + r - sqrt(3/7)*r
    0 = .5 + .345346*r
    -.5 / .345346 = r would mean r is -1.4478.. But the back of the book says otherwise ;-( what did I do wrong here?
     
  5. Jan 30, 2014 #4
    Yo shane, if you are reading this you should email me the title/author of your textbook and the problem number and i'll have a clearer solution posted here tonight
     
  6. Jan 30, 2014 #5
    yo check your email shane
     
  7. Jan 30, 2014 #6

    haruspex

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    Right.
    Wrong.
     
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