Posted by AfterLife on Sunday, August 28, 2011 at 9:52pm.
You are almost right, however, it is easier for me to break up the two forces into vertical and horizontal components.
well, first, the force on q from Q is kQq/distance^2 at some angle from the horizonal (arc tan d/x)
the horizontal force is
a) kqq/distanc^2*cosTheta
if you calculate the horizontal force component from -Q, it is the same but in a opposite direction, so the net horizontal force is ZERO.
Vertical forces:
vertical force from Q (top) is
kQq/distance^2 * sinTheta in the down direction
vertical force from -Q (bottom) is the same, and in the down direction.
so the net force is downward,
2kQq/(d^2+x^2) * sin (arctan d/x)
Thank you very much :D
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