# Ranking Electric Potential of Electrons in a E Field

1. Oct 18, 2013

### Dgray101

1. The problem statement, all variables and given/known data

Note this is a PRACTICE midterm problem. Not due for homework or for a grade. My exam is on Tuesday

"an electron moves from point A to point B in a uniform electric field as show below. Rank the electrons in diagrams 1 through 2 by the changes in potential from greatest to least when traveling from A to B"

I can't attach an image but it's VERY straight forward.

--------------------> (Direction of E field) --------------->(Direction of Efield)

Point A Point B Point B Point A

The distances are the same. Just swapped.

2. Relevant equations

V = U/q

3. The attempt at a solution
My reasoning was (A then B) because in the first diagram, we must do work against the electric field, increasing the potential energy of the electron, therefor increasing the electric potential. The converse is true for part B, we move in the direction of the force and so the potential energy of the electron decreases, and so should the electric potential. I am messing up the concept somewhere along the road because the answer is actually B then A.

2. Oct 18, 2013

### Dgray101

I think I have answered the question but I need validity. The reason that the potential from B is more then the potential at A is because Electric Potential is defined independent of a test charge. So as you go with the electric field, regardless of the charge, you go to lower values of electric potential. If you go opposite the electric field, you go to values of HIGHER electric potential regardless of the charge.

3. Oct 19, 2013

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.

Yes, you're reasoning (in post #2) is correct. It's important to remember that electric potential and electric potential energy are not the same thing.

The way I think of it is to remember that E-fields point away from positive and toward negative charge, and the potential is higher around a positive charge and lower around a negative charge. Your reasoning in post #1, if applied to a positive (test) charge, is also a good way to think of it.