[SOLVED] Electric Force 1. The problem statement, all variables and given/known data http://session.masteringphysics.com/problemAsset/1001926/10/knight_Figure_25_46.jpg What is the force F on the 1 nC charge at the bottom? Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures. 2. Relevant equations F = kq1q2 / r^2 a^2 + b^2 = c^2 Sine Law 3. The attempt at a solution As a reference, the "focus" charge is the 1nC at the bottom. Note that this appears to be a semicircle. Due to this fact, I conclude that the distance between the focus charge and the -6nC is also 5cm. Because both 2nC charges have the same charge, distance, and angle with respect to the origin (Focus), I conclude that they cancel each other out along the X axis. They will both push the focus downwards, however there should be no X axis movement. Since the -6nC charge is along the Y axis, I conclude that this charge will not affect the focus' net X force either. Therefore, F_x = 0 The force given between the 2nC on the left, and the focus, is F = kq1q2 / r^2 Where: k = 9*10^9 q1 = 1*10^-9 (Focus) q2 = 2*10^-9 (Leftmost charge) r = 5cm = 0.05m F = 0.0000072 Due to Sine Law, the X and Y components of this force are equivalent, since the angle is at 45 degrees. Sqrt(X^2 + Y^2) = F^2 Sqrt(2Y^2) = F^2 2Y^2 = F^4 Y^2 = (F^4)/2 Y = Sqrt((F^4)/2) Y = 3.6656415536710623664939771731515 * 10^-11 Since the rightmost charge is identical with respect to the focus, it will put out the same force along the Y axis. Now to include the -6nC charge. F = kq1q2 / r^2 Where: k = 9*10^9 q1 = 1*10^-9 q2 = -6*10^-9 r = 0.05 F = -0.0000216 So, the net force on the focus should be: = "-6nC" + "2nC:Y" + "2nC:Y" = "-6nC" + 2*"2nC:Y" = -0.0000216 + 2*3.6656415536710623664939771731515 * 10^-11 = -0.00002159992668716892657875268 Rounded to 3 significant digits is 2.16*10^-5 Expressing answer as asked: 0,2.16*10^-5 This answer is incorrect. My mistake is where?