Potential and potential energy

[Sigma Rho]

An apparently easy question that has got me baffled! The question is simply "What is the difference between the potential of a body and the potential energy of a body, and why is it sometimes more useful to consider the potential rather than the potential energy?".

I know the answers (I think! :smile: ), but I'm having great difficulty in putting them into words. I guess there would be no marks for just copying stuff out of the text book, so how do I go about explaining it?

I'm not very good at "wordy" questions (I much prefer to work with equations), so if anyone has any advice about this kind of question in general, I'd be glad of it.

Thank you.

[e(ho0n3]

From my understanding, potential energy is the energy associated with the ability to do work.

I've never heard of a body having potential before. Mathematically speaking, whenever a force field (vector field) is conservative, there exists a scalar potential function f such that \vec{F} = \nabla f. I don't know if this has anything to do with the potential of a body though.

[turin]

I will say "voltage" instead of "potential" for clarity (voltage = potential).

Potential energy comes from the interaction between two objects (i.e. two charges).
Voltage is bascially the potential energy stripped of one of the objects, so it's even more abstract.

Let's take charge as an example. It interacts electrically. You can sort of think of the voltage at a point in space as what the potential energy would be if there were a unit charge at that point. This makes some calculations more convenient, because then you can just scale the voltage by the amount of charge that is there to get the potential energy. One key difference between voltage and potential energy is that, while voltage exists (is defined) at every point in space, potential energy doesn't exist at any point in space. Furthermore (to amplify the distinction), voltage only requires one source charge (that is not necessarily interacting with any other charge), whereas potential energy requires two charges (that interact with each other).

The distinction is analogous to the distinction between force and force field. For instance, the Coulomb force is the force that one charge exerts on another. The electric field is the force that a charge would exert on a unit charge if there were a unit charge to exert the force on, but there does not have to be another charge out there for there to be an electric field out there. There does have to be a charge out there for there to be a Coulomb force out there. Thus, in this analogy, the coulomb force is analogous to the potential energy, and the electric field is analogous to the voltage.

You can observe these relationships:

F = qE
U = qV
F = -gradU
E = -gradV

or there reciprocals:

E = F/q
V = U/q
U <- integration of F.dx
V <- integration of E.dx

[e(ho0n3]

It seems that in physics there is a negative sign involved when determining the potential, i.e. \vec{F} = - \nabla f. Why is this?

[e(ho0n3]

The distinction is analogous to the distinction between force and force field.
I was just thinking this. Very nice analogy turin. I think this should clear things up.

[turin]

It seems that in physics there is a negative sign involved when determining the potential, i.e. \vec{F} = - \nabla f. Why is this?It is more of a matter of convention than anything else, AFAIK. In my unofficial opinion, though, I think it speaks something of the 2nd law of thermodynamics.

[robphy]

"Potential" is a property of a conservative electric field.
"Potential energy" is a property of a point charge in that electric field.

[robphy]

Consider a one-dimensional potential-energy diagram.
Consider a small displacement near a stable equilibrium point.
The restoring force is minus the slope of the potential energy curve.
When displaced to the left, the slope is negative. So, the restoring force is positive ("to the right").
When displaced to the right, the slope is positive. So, the restoring force is negative ("to the left").

[Gza]

It seems that in physics there is a negative sign involved when determining the potential, i.e. \vec{F} = - \nabla f
. Why is this?

The negative sign arises from the very definition of potential energy; the one you learned in mechanics. Remember that the potential energy of a system is the work done at no change in kinetic energy in a conservative field (in this case the electric field). This quantity is of course divided through by the second charge in the system, but nonetheless: Imagine a fixed positive charge in space. Now imagine a positive "test" charge being brought closer. This charge isn't accelerating either, it's coming in at no change in kinetic energy (finding the potential.) Now thinking of a free body diagram, you should see an electric force vector pointing out of the test charge parallel and in the same direction as the electric field of the source charge. In order to keep the charge from accelerating (and thus changing its kinetic energy), one must "push" forward on the test charge with a force the same magnitude as the electric force vector caused by the source. Only problem is that this vector is pushing in the wrong direction, so you simply multiply it by a negative one; now you have a vector of the correct magnitude and direction as to cause no acceleration of the test charge. Now simply integrate over the distance and you have the potential.