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Attraction of -Q to +Q in x and y directions

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the magnitude and direction of the force on the charge -Q on the figure in the attached document.

    2. Relevant equations

    F = k(q)(Q)/r2r-hat

    3. The attempt at a solution

    So I have a force in the x direction and a force in the y direction.

    Fx = k(Q1)(-Q)/(a) x-hat

    Fy = (k)(Q2)(-Q)/(a) y-hat

    F = Fx + Fy = -kQ22/(a)2 x-hat + -kQ12/a2 y-hat

    Is this right? Or can more be done to solve it?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Oct 14, 2013 #2

    Simon Bridge

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    Almost there - check the directions of the forces.
    Your formula has the -Q charge being pushed in the -x direction.
    Is that what would happen?

    In general - the force experienced by charge q due to charge Q is:
    $$\vec{F}=\frac{kQ}{r^3}q\vec{r}$$ Here r points from Q to q.
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