A proton moves in a constant electric field
from point A to point B.The magnitude of the electric field is 6.4 × 104 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 0.50 m, what is the change in the proton's electric potential energy, EPEA - EPEB? 
A) -2.4 × 10-15 J
B) -3.2 × 10-15 J
C) +1.2 × 10-15 J
D) -5.1 × 10-15 J
E) -1.8 × 10-15 J
Correct Answer:
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