1. The problem statement, all variables and given/known data A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +9.0 10-4 J. (a) Find the magnitude and direction of the electric force that acts on the particle. _______N the direction of motion (along, against, perpendicular) (b) Find the magnitude and direction of the electric field that the particle experiences. _______N/C the direction of motion (along, against, perpendicular) 2. Relevant equations PE = qV C = q/V E = F/q V = Ed 3. The attempt at a solution I have used the four formulas above to reach two answers. I'm no good with LaTeX so, i apologize but bear with me. PE = qV 9e-4 J = (1.5e-6 C)V V = 600 V C = q/V C = (1.5e-6)/600 C = 2.5e-9 F V = Ed 600 = (.15)E E = 4000 N/C E = F/q 4000 = F/(1.5e-6) F = .006 N As for the direction...I believe the force is along the direction of motion...and the E field is perpendicular? I know the formulas and am doing the work right but if someone could clarify if I have I am understanding it right and using the right numbers I would appreciate it greatly. Thank you.