PHYSICS(HELP)

Eight charges (50μC each) are arranged on the corners of a cube side length 1 cm. What is the magnitude of the force on each charge in Newtons?

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  1. each F = k q^2/d^2
    side of cube length s = .01 meter
    now lets put one corner at the origin and look at the one up and across
    d = sqrt [ (s sqrt 2)^2 + s^2]
    = s sqrt(3)
    and d^2 = 3 s^2
    now you need directions i,j,k components
    first
    component in x y plane, cosines of angles
    cos Angle = sqrt 3/sqrt 2
    now that is split into x and y
    i component sqrt 3/sqrt 2/sqrt 2 = sqrt3/2
    j same as i
    now the vertical (k) component is 1/sqrt 3 F
    so force due to the one at the origin
    = (k q^2/3 s^2)(sqrt3 /2)i + 1/sqrt3 j + 1/sqrt3 k)

    you have to go through that geometry for the effect of each charge (seven of them) on that one opposite and up and sum them. I think I did the hardest corner though.

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    posted by Damon
  2. 700000

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