A charge 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

Question

05-12-2018
Physics

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A charge 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

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joshnajd joshnajd · Answer 1 · 2019-09-26T13:32:58+02:00

Answer:

The correct answer is not present in the options given. The right answer is [tex]7.8 \times 102 \ J/C[/tex] as the unit of potential energy difference is [tex]J/C.[/tex]

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

[tex]E=V/d[/tex]

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

[tex]E= \Delta V/d[/tex]

Thus,

[tex]\Delta V= Ed = 6.5 \times 10^4 \times 1.2 \times 10^-^2[/tex]

[tex]\Delta V= 7.8 \times 10^2 \ J/C[/tex]

So the correct answer is not present in the options given. The right answer is [tex]7.8 \times 102 \ J/C[/tex] as the unit of potential energy difference is J/C.