A proton moves in a constant electric field E from A to pointB. The magnitude of the elctric field is 4.2x10^4 n/c; Thedirection opposie to that of the proton. If the distance from pointA to point B is 0.18m, what is the change in the proton's electricpotential energy, EPEA-EPEB?

Answer 1

Force (retarding) is F = Q E = 1.6*10^-19 * 4.2*10^4 Newtons
It is losing kinetic energy and gaining potential energy
Work = F d = F * .18 Joules