Question

What is the magnitude of the difference in potential energy for a charge of magnitude 6.4×10^−19 coulombs moving from point A to point B in an electric field of 6.5×10^−4 newtons, given the distance between A and B is 1.2×10^−2 meters and the path between A and B is parallel to the field?

A) 4.16×10^−22
B) 4.94×10^−22
C) 5.42×10^−22
D) 6.18×10^−22

Answer 1

The magnitude of the difference in potential energy is 4.16×10^-22 joules.

Step-by-step explanation:

The potential energy of a charge in an electric field can be calculated using the formula:

Potential Energy = Charge * Electric Field * Distance

In this case, the magnitude of the charge is 6.4×10-19 coulombs, the magnitude of the electric field is 6.5×10-4 newtons, and the distance is 1.2×10-2 meters.

By substituting these values into the formula, we get:

Potential Energy = (6.4×10-19 C) * (6.5×10-4 N/C) * (1.2×10-2 m) = 4.16×10-22 joules.

The magnitude of the difference in potential energy is 4.16×10-22 joules, so the correct answer is A) 4.16×10-22.