1. The problem statement, all variables and given/known data Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = -0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges? 2. Relevant equations Columbs law (force between two point charges) F = 1/4 piEo times abs value of q1*q2 divided by r^2 3. The attempt at a solution first i have to find the force that Q exerts on q1 (F1 on q1) i use coulumbs law F = k * [q1q2]/[r^2] k = 8.988 e 9 so i plug in q1 and Q and the distance between them and i get.... F1 on q1 = .29 i break that down into components x and y F1x= .29cos(.40/.50) = .287 F1y = .29sin(.30/.50) = .00301 so now i find the force that q2 exerted on q1 (F2 on q1) q2 and q1 are both on the vertical axis so there will only be a y component so i use cuolumbs law again and now i plug in q1 and q2 and r is the distance between them so.. F2= k [2.0 e -6][2.0 e -6] / (.60 ^2) = .1 so now i add the x components and y components up there is only one x component and 2 y components sooo F1x = .287 F1y + F2y = .10301 F = square root of ( .287^2 + .10301^2) which is equal to .30 (2 sig figs) then to find the angle i would use tan(theta) = Fy/Fx which gives me an angle of 19.7 degrees It is supposed to be in the +x direction so i think i am supposed to subtract it from 180 and that should give me my angle im afraid to type the answer in into masterinphysics because i have only a few attempts left...so i want to make sure im right before i do anything!