Acharge q of magnitude 6.4 × 10-19 coulombs moves from point a to point b in an electric field of 6.5 × 104 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 the 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

The correct answer is not present in the options given. The right answer is as the unit of potential energy difference is

Explanation:

The electric potential energy is the amount of work done to move a point charge from one point to another. The difference in the potential energy can be derived as the change in the potential energy in different points.

As it is known that potential energy is directly proportional to the electric field in which the point charge exists, the representation will be

Here, E is the electric field, V is the potential energy and d is the distance of the point charge to be moved.

So, if we need to move a charge q from point A to B, the change in potential energy will be

Thus,

So the correct answer is not present in the options given. The right answer is as the unit of potential energy difference is J/C.