A uniform electric field,with a magnitude of 600 N/C,is directed parallel to the positive x-axis.If the potential at x = 3.0 m is 1 000 V,what is the change in potential energy of a proton as it moves from x = 3.0 m to x = 1.0 m? (qp = 1.6 x 10-19 C)
A) 8.0 x 10-17 J
B) 1.9 x 10-16 J
C) 0.80 x 10-21 J
D) 500 J
Correct Answer:
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