Hi guys I came upon this problem while solving in a book, there is something I am confused about the way he solved it in the answers. (II)Three charged particles are placed at the corners of an equilateral triangle of side 1.2m. The charges are 4uC,-8uC,and -6uC. Calculate the magnitude and direction of the net force on each due to the other two. The charge -4uC is on top and -6.00uC is on the right and -8uC is on the left I answered this by calculating the magnitude of each charge then getting the net charge by adding the vector components of each. F12 = 0.2N; F13 = 0.15N; F32 = 0.3N; F31 = 0.15; F23 = 0.3N; F21 = 0.2N; I concluded the force on charge Q1 should be pointing to Q2 since its larger magnitude so I first calculated the components vectors F12 as negative as so:- F12x = -cos(60)*0.2 = -0.1N; F12y = -sin(60) * 0.2) = (-(3^1/2) / 10) N F13x = cos(60)(0.15) = (3/40)N F13y = sin(60)(0.15) = 0.129N Now since the force is favored to go to F12 then we can add F12 to F13 we are subtracting anyway thats why I did F13 as positive,since I know that force net should go to F12; F1x = -0.3 + 3/40 = -1/40N F1y = -(3^1/2) / 10 + 0.129 = -0.044N But the way he answered in the book is really different. the book answer is as follows: F1x = -(0.1997N)cos(60) + (0.1498)cos60 = -2.495 * 10^-2; F1y = -(0.1997)sin(60) - (0.1498N)sin60 = -3.027 * 10^-1; Notice that he subtracted the y component vector which really doesn't make sense I mean both are attracting the positive proton so it wouldn't make sense to subtract. Did I miss something here? I also notice that 0.1997 F12 is really close just my rounding must have been incorrect. I also didn't include here the calculation for the force relative to point 2,since if I know what I did wrong here same thing apply to point 2 calculation(thats if I did something wrong). Thanks alot.