F = k × Q1 × Q2 / r² where F = force (in Newton) k = constant of coulomb = 9E9 Nm²/C² Q1 = first charge (in Coulomb) Q2 = second charge (in Coulomb) r = distance between centers of charges (in meter) -> force between Q1 and Q3 (= Fa) Fa = (9E9) × (5.65E-8) × (2.63E-8) / (0.323²) Fa = 1.2819E-4 N Q1 and Q3 have the same sign, so the direction of Fa is straight upward, and Fax = 0 N Fay = 1.2819E-4 N -> force between Q2 and Q3 (= Fb) First find the distance between the charges r = v( (0.189)² + (0.323)² ) = 0.374 m Fb = (9E9) × (-1.93E-8) × (2.63E-8) / (0.374²) Fb = -3.2619E-5 N Q2 and Q3 have different signs, so the direction of Fb is from Q3 to Q2. The angle of Fb will be: ? = arctan( -0.323 / 0.189 ) = -59.67° and Fbx = 3.2619E-5 × cos(-59.67) = 1.6474E-5 N Fby = 3.2619E-5 × sin(-59.67) = -2.8153E-5 N -> find the x- and y-component of the resulting force (=Fr) Frx = Fax + Fbx = 0 + 1.6474E-5 = 1.6474E-5 N Fry = Fay + Fby = 1.2819E-4 - 2.8153E-5 = 1E-4 N -> find the magnitude of Fr Fr = v( (1.6474E-5)² + (1E-4)² ) = 1.014E-4 N find the angle of Fr f = arctan( 1E-4 / 1.6474E-5 ) = 80.6°