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Point charges acting on a point

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Three point charges are located at the vertices of an equilateral triangle. The charge at the top vertex of the triangle is -4.4 µC. The two charges q at the bottom vertices of the triangle are equal. A fourth charge 2 µC is placed below the triangle on its symmetryaxis, and experiences a zero net force from the other three charges, as shown in the figure below. Find q.
    One side of the triangle is 7.3m. The distance from 2 µC to the triangle is 4.8m.

    -4.4 µC {1}
    2 µC {2}

    2. Relevant equations

    F=kQ1Q2/r^2
    E=kQ/r^2


    3. The attempt at a solution

    F (1 on q)= k (-4.4*10^-6)q/7.3^2

    (7.3/2)^2+(4.8)^2=b^2
    =6.03

    F(2 on q)=k(2*10^-6)q/6.03^2
     
  2. jcsd
  3. Sep 5, 2011 #2

    vela

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    You actually want to calculate the forces on the 2 uC charge, not on the q's, since the problem tells you the net force on that charge is 0.
     
  4. Sep 5, 2011 #3
    So something like this?

    F (1 on 2)= k (4.4*10^-)(2*10^-6)/(11.12)^2

    F(q on 2)= k*Q*(2*10^-6)/(6.03)^2
     
  5. Sep 5, 2011 #4

    vela

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    Yes, and just remember you need to sum them as vectors.
     
  6. Sep 5, 2011 #5
    So F(1 on 2)x=0
    F(1 on 2) y=6.305*10^-4

    but how do I apply this to F(q on 2)

    F(q on 2)x= F(q on 2) Cos 60
    F(q on 2)y= F(q on 2) Sin 60

    but F(q on 2)= k*Q*(2*10^-6)/(6.03)^2

    and Q is unknown...

    also would 2F(q on 2)+F(1 on 2)=0
     
  7. Sep 5, 2011 #6

    vela

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    Pretty close. I think you have the right idea.

    For simplicity, let's assume q is positive. The force of charge 1 on charge 2 will point up. The force of a q-charge on charge 2 will point away from the q. So in component form, you have
    \begin{align*}
    \sum F_x &= F_{q~\textrm{on}~2} \cos 60 - F_{q~\textrm{on}~2} \cos 60 = 0 \\
    \sum F_y &= F_{1~\textrm{on}~2} - 2F_{q~\textrm{on}~2} \sin 60 = 0
    \end{align*}
    The first equation simply tells you the x-components cancel out, regardless of what q is, which you could have deduced beforehand based on symmetry. The second equation is the one that will let you solve for q.
     
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