An electron in the atmosphere is moved upward through displacement d by an electrostatic force due to an electric field
W done on the electron by the electric field (tlU = – W) gives the relation. A second Key Idea is that the work done by a constant force F on a particle undergoing a displacement d is W= F·d.Finally, the third Key Idea is that the electrostatic force and the electric field are related by F = q£, where here q is the charge of an electron (= -1.6 X 10-19 C). Substituting for Fin taking the dot product yield IV = qE’ d = qEd cos (),where () is the angle between the directions of £ and d. The field£is directed downward and the displacement d is d;erected upward so ()= 180·. Substituting this and other data into we find W = (-1.6 X 10-19 C)( 150 N/C)(520 m) cos 180″ = 1.2 X 10-14 J.tlU = -IV = -1.2 X 10’14 J.This result tells us that during the 520 m ascent. the electric potential energy of the electron decreases by X 10-14 J. In the figure. a proton moves from point Ito point j in a Uniform electric field directed as shown. (a) Does the electric field do positive or neg-native work on the proton (b) Does the electric potential. l energy of he proton increase or decrease.