1. The problem statement, all variables and given/known data Given the following information: Givens: A = +2.0mC B = +4.0 mC C = - 4.0mC All of the sides of the triangle are 1.0 m The inside angles are 60° 1. Determine the force acting on charge C as a result of A and B. Be sure to include a direction. 2. Determine the electric field strength and direction at a point half way between B and C. IGNORE A in this question. 3. Determine the total electric potential energy of the system of charges. Be sure to include all possible charge interactions. 4. Determine the total electric potential as a result of charges B and C at a point halfway between them. Ignore A in your thinking. 2. The attempt at a solution 1. Fb = kQq/d^2 Fb = (9x10^9) (4x10^-6)(4x10^-6) / 1^2 Fb = 0.144 N Fa = kQq/d^2 Fa = (9x10^9) (4x10^-6) (2x10^-6) Fa = 0.072 N x = 0.144 + 0.072 cos 60 y = 0.072 sin 60 Therefore, resultant = 0.079 N[19 degree above horizontal] 2. Eb = kq / d^2 Eb = (9x10^9)(4x10^-6) / 0.5^2 Eb = 144000 N/C Ec = kq / d^2 Ec = (9x10^9)(4x10^-6) / 0.5^2 Ec = 144000 N/C E = Eb + Ec E = 144000 + 144000 E = 288000 N/C Therefore, electric field strength is 2.9x10^5 N/C 3. Pe = k(qa qb / rab + qa qc / rac + qb qc / rbc) Pe = (9x10^9)(2x10^-6 * 4x10^-6 / 1^2 + 2x10^-6 * 4x10^-6 / 1^2 + 4x10^-6 * 4x10^-6 / 1^2) Pe = (9x10^9) (8x10^-12 + 8x10^-12 + 1.6x10^-11) Pe = 9.216 x 10^-24 J Therefore, total electric potential energy of the system of charges is 9.2 x 10^-24 J 4. Needs help solving this question. Thanks!