Coulomb’s Forces


Consider the three charges in the figure below, in which d = 4.7 cm, q = 11 nC, and the positive x-axis points to the right. What is the force Fvec on the 5 nC charge? Give your answer as a magnitude and a direction.

2. Relevant equations

F= (K*Q*q)/r^2

3. The attempt at a solution

First I found the distance between +5 and 11.
(.047^2)+(.03^2)=.055758 m

arctan(.047/.03)= 57.45

Then the magnitude: .000164

The the magnitude of +5, -5 –> .00025 (Horizontal component)

cos 57.45 = x/.000164 = 8.823X10^-5 (Vertical component of line)

sin 57.45 = x/.000164= 1.3824×10^-4 (Horizontal component of line)

Add horizontal components and use that and vertical to make a triangle and find the third side.

So my answer is incorrect, can someone help me out please. Thank you.

Attached Images
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