Sign on change in electric potential energy

[henry3369]

Homework Statement

I'm trying to figure out why potential energy decreases in this situation
http://imgur.com/9PfjRVE

Homework Equations

ΔU = FΔy

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.

 

[ehild]

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

 

[henry3369]

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

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.

 

[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.

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

 

[henry3369]

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

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?

 

[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.

 

[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?

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

 

[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 y_A to y_B is W=qE(y_A-y_B= 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.

 

[ehild]

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

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