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Homework Help: What is the magnitude of each fixed charge in coulombs?

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Four point charges, q, are fixed to the four corners of a square that is 13.1 cm on a side. An electron is suspended above a point at which its weight is balanced by the electrostatic force due to the four point charges, at a distance of 20 nm above the center of the square. (The square is horizontally flat, and the electron is suspended 20 nm vertically above the center of the square.) What is the magnitude of each fixed charge in coulombs?
    ___ C

    What is the magnitude of each fixed charge as a multiple of the electron's charge?
    ___ e

    2. Relevant equations

    F = (k*q1*q2)/d^2

    3. The attempt at a solution
    Since there is a net force of 0 for the electron, I assume that all q-charges are positive since they pull with equal force from all four directions. Since this is a 3-dimensional problem I use vectors when I calculate the forces on e(Fnet(x) = 0 , Fnet(y) = 0). When I try to break the forces on e up into vectors the angle that I get is 90 degrees. This somewhat makes sense since the distance of 20 nm is so small it can almost be negligible. I am not sure though, where to go from here.
  2. jcsd
  3. Jan 26, 2012 #2


    Staff: Mentor

    so it looks like you have a pyramid with gravity pulling the suspended electron down and 4 symmetrically placed charges at each corner of the base pushing it up. I don't see where the 90 degrees comes in.

    Because of the symmetry the horizontal components of the four charges cancel out leaving only the vertical. You know how to calculate the force on the suspended charge from these 4 and you know how to calculate the force of gravity on it right.
  4. Jan 26, 2012 #3

    4*(k*q*e)/d - Me*g = 0
    where d is the horizontal distance from a to e.
    Am I correct here?
  5. Jan 26, 2012 #4
    4*(k*q*e)*sin(theta)/d^2 - Me*g = 0
  6. Jan 26, 2012 #5


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    Staff Emeritus
    Science Advisor
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    Consider the following right triangle. One leg is along the diagonal of the square, from one corner to the center of the square. The length of this leg is [itex]\displaystyle(\sqrt{2})\cdot6.55\ \text{ cm}[/itex]. The other leg is vertical, from the center of the square, up to the electron. The length of this leg is 20 nm = 20×10-7 cm.

    The very small angle at the vertex of the triangle located at a corner of the square, has a measure of about [itex]\displaystyle \theta=\frac{20\times10^{-7}}{6.55}\approx 3.0534351\times10^{-7}\text{ radians}\approx 1.74949^\circ\times10^{-5}\,.[/itex] Thus the other acute angle, the one with its vertex at the location of the electron has a measure of ≈ 89.9999825051°. That's temptingly close to 90°, but don't round it off.

    The vertical component of the Coulomb force is a tiny fraction of the total Coulomb force.
  7. Jan 26, 2012 #6
    Thank you both for your help I figured it out.
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