1. The problem statement, all variables and given/known data Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2? 2. Relevant equations Fcos(θ) = Fx Fsin(θ) = Fy θ = Tanh(Fy/Fx) F = (Fx^2 + Fy^2)^1/2 F(2on3) = K(q1q2)/r^2) 3. The attempt at a solution Using the program's help function, I have found the components (F1on3)x = 2.96x10^-4 (F1on3)y = 5.13x10^-4 So the only thing left was to find F(2on3) in the same way I found F(1on3). Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this: Fy = 0 (sin0=0) Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2 = 1.19x10^-3 This makes sense given that it should attract. Here's where the problem falls off of the rails... I need to find Theta So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006 I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?