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?

Answer 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.

Answer 2

700000