1. The problem statement, all variables and given/known data How much energy is need to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm? 2. Relevant equations E = kQ/r^2 3. The attempt at a solution I'm really not sure if what I did is close to correct, but I got an answer. 1) I calculated the Force experienced by 1 charge by another (arbitrarily picked): F = k Q^2/d^2 = 3.6 x 10^8 Therefore, if I picked the bottom left charge of the square, it is experiencing a repulsive force in the -y and -x directions each of 3.6 x 10^8 N. 2) The third charge, the upper right, was also taken into account after finding the distance from the bottom left to the top right: F = k q^2/ r^2 = 1.8 x 10^6 N Therefore, the top right charge is repelling the bottom right charge by 1.8 x 10^6 N at a 45 degree angle south of west. 3) I added up the vectors to find a resultant vector of the force acting on the bottom left charge: (tr = top right) a) F(tr)x = 1.8 x 10^6 (cos 45) = 1.27 x 10^6 N (in the negative x direction) b) F(tr)y = 1.8 x 10^6 (sin 45) = 1.27 x 10^6 N (in the negative y direction) The resultant vector consists of: F(x) = (1.27 x 10^6) + (3.6 x 10^8) = 3.61 x 10^8 (negative x direction) F(y) = 3.61 x 10^8 (negative y direction) Resultant force = sqrt [(3.61 x 10^8)^2 + (3.61 x 10^8)^2] = 5.1 x 10^8 N (at 45 degrees south of west) I'm not sure where to go from here. I'm wondering if I should have calculated work from each of the seperate charges acting on the arbitrarily picked bottom left charge and added them up?: a) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J + b) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J + c) W= Fd = (1.8 x 10^6 N) x (0.035m) = 6.3 x 10^4 J Total Energy = 1.81 x 10^7 J Then the TOTAL ENERGY for all 4 charges should be 1.81 x 10^7 multiplied by 4? (Because there are 4 charges): = 7.23 x 10^7 J[/B] HELP PLEASE! TEST ON MONDAY!