Acharge q of magnitude 6.4 × 10^-19 coulombs moves from point a to point b in an electric field of 6.5 × 10^4 newtons/coulomb. the distance between a and b is 1.2 × 10^-2 meters and the path between a and b is parallel to the field. what is the magnitude of the difference in potential energy?

a. 1.2 × 10^-15 joules
b. 2.3 × 10^-15 joules
c. 3.2 × 10^-15 joules
d. 6.4 × 10^-15 joules
e. 6.4 × 10^-14 joules

In this problem we have the electric field intensity E:

E = 6.5 × newtons/coulomb

We have the magnitude of the load:

q = 6.4 × coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × )(6.5 × )(1.2 × )

PE = 5.0 x joules

None of the options shown is correct.

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