A charge 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 the potential energy?

Question

dboome · Answer

Well, since it sounds like the field is constant (uniform) it shouldn't quite matter where you are in it; the charge's potential energy will be the same no matter where it moves so long as the electric field's magnitude does not change. There is, however, a negative amount of work done by the electric field when it moves from A to B (negative because the charge is being repelled towards B, which is described as being parallel to the electric field from A) that is equal to the force applied to the charge multiplied by the distance traveled, if that is of concern.

Think about the gravity of the earth, as gravity and electo-magnitism are very analogous. It doesn't matter where are on the surface of earth, you feel the same magnitude of gravity's pull (so long as you don't increase or decrease elevation --> uniform gravitational field) and since you're assumingly the same mass (analogous to the charge of q), you will have the same potential energy. E.g. 2 rocks of equal mass on a level table 10 meters apart have equal potential energy. You're charge is still feeling the same 6.5 × 10^4 newtons/coulomb force at B that it was at A, and since its charge is the same that implies its potential energy is the same.

irishboy123 · Answer

$W = F * s = E q * s = \Delta U$

gravity analogy is good!