# Homework Help: Sign on change in electric potential energy

1. Apr 15, 2015

### henry3369

1. The problem statement, all variables and given/known data
I'm trying to figure out why potential energy decreases in this situation
http://imgur.com/9PfjRVE [Broken]

2. Relevant equations
ΔU = FΔy

3. The attempt at a solution
If I define up to be positive, and ΔU = FΔy. In this picture, force is pointing down (so it is negative) and the height decreases so Δy is negative.
Then ΔU = (-F)(-Δy) = FΔy. So then change in electric potential energy is positive, which means it should be increasing. But in the picture it says that U decreases.

Last edited by a moderator: May 7, 2017
2. Apr 15, 2015

### ehild

Force is negative gradient of the potential, so ΔU = - FΔy.

3. Apr 15, 2015

### henry3369

Is there a reason it is negative? Everywhere in my book it keeps defining it as positive. And if it is anything like gravitational potential energy, shouldn't it positive? U = mgh for gravitational PE and U = qEy for electric potential energy.

4. Apr 15, 2015

### davidbenari

Think of this though: With the gravitational potential energy $mgh$ you don't use Final-Initial when you're calculating a work done. You do Initial-Final. Which is effectively $-\Delta U$ If by $\Delta$ we understand a procedure of Final-Initial.

For example: The work done by the gravitational field as an apple falls from a height of 5 m.

$m_{apple} (9.8m/s^2) (5m) - m_{apple} (9.8m/s^2) (0m)$

5. Apr 15, 2015

### henry3369

Now that I think about it, Work = FΔy = -ΔU.

So I'm assuming you can't set reference points anywhere you would like when it comes to electric potential energy? For example, if I made the reference point the height of an electron, and an electron moves below the reference point, I can't automatically assume that potential energy is negative?

6. Apr 15, 2015

### ehild

The potential energy in a point A is defined as the work done by the force when the body moves from A to the place where the potential is zero.

So U(A)-U(0)=FΔr. U(A) is the initial potential energy and U(0) is the final one. We define the change as the final value - initial value. ΔU=U(final)-U(initial) so U(A)-U(0)= -ΔU: ΔU = - FΔr.
In cases of gravity, U = mgh, and the force of the Earth points downward. In case of a falling body, the potential energy decreases.

7. Apr 15, 2015

### henry3369

Adding to this, does the sign of the charge determine whether or not the potential energy is negative at this position?

8. Apr 15, 2015

### ehild

The electric field strength is the force exerted on a unit positive charge, and U is the potential energy of the unit positive charge. The force on a charge q is F=qE, and the work by the field when the charge moves from yA to yB is W=qE(yA-yB= U(A)-U(B)) = - q ΔU.
The charge of the electron is negative. In the electric field, it moves in the direction of decreasing potential energy, which means increasing potential.

9. Apr 15, 2015

### ehild

The potential is given, but the potential energy of the charge depends on the sign of the charge .