ELECTROMAGNETISM PHYS- ELECTRIC POTENTIAL ENERGY**HELPPP!!!!Previous incorrect answers:720.271091.0226.511121.21180.54

particle were to pass through point A, what would be its speed vA at that point? This is also an energy conservation problem, E. = Ef. The initial energy is all ELECTRIC POTENTIAL ENERGY U = kq192/r. The final energy should be a combination of potential and kinetic. Make sure to keep all of the minus signs and be careful about finding the distances between the charges. Charge Q1 = 35.41 mC is placed R = 33.15 cm to the left of charge Q2 = 54.23 mC, as shown in the figure. Both charges are held stationary. Point A is located R3 = 9.945 cm to the right of Q1. A particle with a charge of q = -4.551 JC and a mass of 22 38.31 g is placed at rest at a distance R2 = 29.84 cm above O2. If the particle were to be released from rest, calculating .R its exact path would be a challenging problem. However, it is possible to make some definite predictions about the future motion of the particle If the path of the particle were to pass through point A, what would be its speed va at that point? 27 VA= m/s Incorrect Constants Electrostatics e = 1.60 x 10-19 C Electrostatics F = kq192/12 F = qE 1 eV = 1.60 x 10-19 J E = kq/r2 AU = -W = - [F . ds k= = 8.99 x 109 Nm2/C2 U = kq192/r U = qV 60 = 8.85 x 10-12 C2/ (N . m?) V = kq/r mp = 1.67 x 10-27 kg me = 9.11 x 10-31 kg