posted by Anonymous on .
A proton moves in a constant electric field E
from point A to point B. The magnitude
of the electric field is 4.2 x10^4 N/C; and it is directed as shown in the drawing, the
direction opposite to the motion of the proton. If the distance from point A to point B
is 0.18 m, what is the change in the proton's electric potential energy, EPEA – EPEB?
ΔPE = - W = - q•Δφ = - e•E•Δx =
- 1.6•10^-19•4.2•10^4•0.18 =
= -1.21•10^-15 J