Question: Three charges are located as shown in the figure.
What is the force (magnitude and direction) on the 25 mC
charge? 
Answer:
You have to add the forces as vectors. Both forces on the 25
mC charge have a magnitude given by
F = (9.0·109)(25.0·10-6)(-5.0·10-6)/(5.0)2 = -0.045 N
You have to add them as vectors knowing that that triangle is
equilateral and that q = 30°. The
x-component of the resulting vector is zero but the y-component is
2Fcosq = 2·0.045·cos 30° =
0.078 N downward. 
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