1. The problem statement, all variables and given/known data Three charges, +Q, +Q, and -Q are placed at the vertices of an equilateral triangle of length "s" on a side. Find the magnitude and direction of the force on one of the +Q charges. 2. Relevant equations ForceElectric=(K*q1*q2/d2) 3. The attempt at a solution I've set the triangle up so that the bottom side is parallel to the x-axis. My +Q charges are in both the bottom left (charge 1) and bottom right (charge 3) corners, while the -Q charge (charge 2) is at the top. Charge 3 is the charge that I have chosen to do the calculation for. Force2 on 3 = (2QK/s2) Force2 on 3 = Force1 on 3, so Force1 on 3 = (2QK/s2) x-component: Forcex:2 on 3 = (2QK/s2)*cos(60°) = (QK/s2) y-component: Forcey:2 on 3 = (2QK/s2)*sin(60°) ∑Fx = Force1 on 3 - Forcex:2 on 3 ∑Fx = (2QK/s2) -(QK/s2) ∑Fy = Forcey:2 on 3 ∑Fy = (2QK/s2)*sin(60°) Fnet = ∑Fy + ∑Fx (vector equation) Fnet = √((∑Fy)2 + (∑Fx)2) Simplified this (hopefully) goes to: Fnet = (2QK/s2) This was my final answer for the magnitude, but I'm not even sure if it is correct, or if my procedure is right. I'm also not sure how to find the direction. Any help would be greatly appreciated! Thank-you!