# Definition of potential energy

1. Sep 27, 2015

### gracy

In my book definition of potential energy difference between two points as the work required to be done by an external force in moving without accelerating charge q from one point to another in electric field of any arbitrary charge configuration.I want to know why without accelerating ?

2. Sep 27, 2015

### Staff: Mentor

Because if you accelerate it, you also change the kinetic energy so that the work you do ends up being equal to the potential energy difference plus the change in kinetic energy.

3. Sep 27, 2015

### gracy

I did not understand.

4. Sep 27, 2015

### Staff: Mentor

Consider a charged particle sitting somewhere in an electric field, at rest. Because it is at rest its kinetic energy is zero.
I do some work on it to move it somewhere else.

Case 1: It's at rest when I'm done moving it. This is the no-acceleration case, and the work that I did to move it is the change in the potential energy. There's no change in kinetic energy because it's at rest when I'm done moving it so the kinetic energy is still zero.

Case 2: It's not at rest when I'm done moving it, instead it's moving at speed $v$. This is the case in which the particle did accelerate, and the particle now has kinetic energy $\frac{mv^2}{2}$. The work that I did is the change in the potential energy from moving the particle, PLUS the work that I did to accelerate the particle and increase its kinetic energy.

Last edited: Sep 27, 2015
5. Sep 27, 2015

### gracy

I understood now.
It becomes $W= ΔU+ ΔKE$
But As per definition
$W=ΔU$
That's why no acceleration.

6. Sep 29, 2015

### ehild

Potential energy is attributed to conservative forces. The work of a conservative force on an object does not depend on the path taken. It depends only on the initial and final positions of the object. Force such forces, a potential energy function is defined, which depends on the position only.
Introducing an external force and zero acceleration into the definition causes lot of confusion. A simpler and less confusing definition of potential energy difference UB-UA between two points A and B is the negative work done by the force while the object moves from A to B: UB-UA=-WAB, or the work done by the force is the difference between the initial and final potential energy.
The potential energy at a point A is equal to the work the force does when the object moves from A to the place where the potential is zero. In case of the field of a point charge Q, the potential energy is zero at infinity.

7. Sep 29, 2015

### gracy

But work energy theorem says $W$=$ΔKE$

8. Sep 29, 2015

### HallsofIvy

No, it does not. The work done is the change in kinetic energy, $\Delta KE$, minus the change in potential energy, plus the work done against friction.

(Strictly speaking the "potential energy" only exists for a "conservative" force, in which there is no friction but if you are given a "potential energy" formula, you might have divided the force into a conservative part (as gravitational force), which has a potential energy, and a non-conservative part.)

9. Sep 30, 2015

### gracy

Ok.
As per my understanding
$Wnet$=$ΔKE$
(According to work energy theorem)

$Wconservative$= $-ΔP.E$
$Wnon conservative$=$ΔP.E$+$ΔK.E$
Am I correct?

10. Oct 1, 2015

### jbriggs444

Equations by themselves do little without a definition of the terms in use.

Yes, the work done by a conservative force on an object moving in its field is equal and opposite to the change in potential energy between the starting point and the ending point of the path through which it moves.

However, the work done by a non-conservative force on an object moving in its field can have nothing whatsoever to do with the change in potential energy between the starting point and the ending point for the simple reason that there is no such thing as potential energy in a non-conservative field.

11. Oct 1, 2015

### gracy

Is it every time applicable
$Wconservative$=$-ΔP.E$=$ΔKE$

Last edited: Oct 1, 2015
12. Oct 1, 2015

### jbriggs444

Again, I urge you to define your terms before simply writing down equations. What does "KE" in the above equation denote? Be specific. Do not simply say "Kinetic Energy". The kinetic energy of what?

13. Oct 1, 2015

### gracy

Depends on context.That's what I am asking is this applicable always?That is irrespective of any scenario or situations?

14. Oct 1, 2015

### gracy

For example During a free fall there is change in KE of the body in the expense of PE.

15. Oct 1, 2015

### nasu

No. This will work only if there are only conservative forces doing the work. So that W is the net work.

16. Oct 1, 2015

### jbriggs444

Again, I urge you not to reply twice to the same posting. And again, I ask you to define the terms in your equation. "It depends" is not a definition.

What I am looking for is something that relates "Wconservative" to "-P.E." and to "KE"

Is KE, for instance supposed to mean the change in kinetic energy of an object subject [only] to a particular conservative force as it moves between a pair of endpoints? And is P.E. supposed to mean the change in potential energy as that object moves between the same endpoints? And is Wconservative supposed to denote the work done by the conservative force on that object over some path between those same endpoints?

17. Oct 1, 2015

### gracy

In my OP there is only conservative force but it says without acceleration that means $ΔKE$=0
But $ΔP.E$ has nonzero magnitude which is equal to work done by conservative (electric)force
So even though it is only conservative force the below equation is not applicable
$W$=$-ΔP.E$ =$ΔKE$
Right?

18. Oct 1, 2015

### jbriggs444

In your original post there are two forces. There is the force from the field whose potential is being evaluated. And there is an external force.

19. Oct 1, 2015

### gracy

Yes
Yes.
Yes.
According to me.You decide this is the
question

A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent in doing so is 100J.

20. Oct 1, 2015

### Staff: Mentor

OK. Now, what is Wnet also equal to? You're lacking one step necessary to get to your last equation below.

21. Oct 1, 2015

### Staff: Mentor

I don't know. Personally, I wouldn't look at it this way. I would let the math resolve the issue for me. Here's a one dimensional example. Suppose I have a charged particle of mass m and charge q in an electric field E (directed along the positive x axis), and I exert a force F along the x axis to move the charge from x1 to x2. If a write a force balance on the system, I get:
$$m\frac{dv}{dt}=F+qE$$
Since an electric field is conservative, it can be represented as minus the derivative of the potential U. So, I have:
$$m\frac{dv}{dt}=F-q\frac{dU}{dx}$$
If I substitute dt=dx/v into this equation, I get
$$mv\frac{dv}{dx}=F-q\frac{dU}{dx}$$
Integrating this equation from x1 to x2 yields
$$\left(m\frac{v^2}{2}\right)_{x_2}-\left(m\frac{v^2}{2}\right)_{x_1}=\int_{x_1}^{x_2}Fdx-q(U(x_2)-U(x_1))$$
The terms on the left hand side of this equation is the change in kinetic energy. The first term on the right hand side is the external work done on the charge. The second term on the right hand side is minus the change in potential energy. So
$$Δ(KE)=W-Δ(PE)$$or
$$W=Δ(PE)+Δ(KE)$$
If the work is done in such a way that the change in kinetic energy is zero, then the work is equal to the change in potential energy.

Chet

22. Oct 1, 2015

### gracy

$Wnet$=$Wconservative$+$Wnonconservative$

23. Oct 1, 2015

### Staff: Mentor

OK, so with that step, and your first two equations, you should be able to verify that your final equation is correct, with a tiny bit of algebra.

24. Oct 1, 2015

### gracy

Oh,yes.Please look at my post 19

25. Oct 1, 2015

### gracy

Are all my equations correct?