1. The problem statement, all variables and given/known data 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= 1.3824x10^-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.