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Equilibrium of Charges using Coloumbs Law

  1. Aug 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge of -0.5 C is located at the origin. A second point charge of 12 C is at x = 1 m, y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium.


    2. Relevant equations
    F=k*((q_1*q_2)/r^2)



    3. The attempt at a solution
    I know that I will have to find the two forces that the point charges will have on the electron at some arbitrary point that we will eventually solve for. We then set those two forces equal to each other and then solve for the distance. Once we have the distance we can easily solve for the components using simple trig... I must be having a problem setting up my equations any help?
     
  2. jcsd
  3. Aug 27, 2009 #2

    kuruman

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    Start by showing what you have done so far. Your strategy is OK, can you do what you said you should do and write some equations down? Also drawing a picture with where you think the electron might be located should help.
     
  4. Aug 27, 2009 #3

    tiny-tim

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    Hi gc33550! :smile:

    I'm not sure whether you've noticed that this is a one-dimensional problem …

    then just use the usual 1/r2 formula, using a different r for each charge, of course. :wink:
     
  5. Aug 27, 2009 #4
    Thanks for your Input. Yes I realize it is one dimensional so hear is what I have so far:
    (I will define 1 as the negative charge and 2 as the positive and e will be the electron)

    F_1e=k*((-.5muC*e)/R_1e^2)
    F_2e=k*((12muC*e)/R_2e^2)

    to find equilibrium:

    F_1e= F_2e

    I am then left with:

    ((-.5muC)/R_1e^2)=((12muC)/R_2e^2)

    I know the distance of R1 to R2 because of trig. So then R_2e=(R_1e+1.12m)
    As far as where the electron is it would be in the third quadrant along a straight line with the other two point charges I know it will have to be close to the negative point charge to balance out the great attractive force of the positive point charge.
    Once again thanks to both of you.
     
  6. Aug 27, 2009 #5

    tiny-tim

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    Hi gc33550! :smile:

    (have a mu: µ and try using the X2 and X2 tags just above the Reply box :wink:)
    (your equations are almost unreadable, but I think I follow that bit :redface: …)

    ok, now substitute for R2 in the first equation, put everything on top, and you have a quadratic equation in R1 :smile:
     
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