# B The total potential energy of two charges Wikipedia mistake?

Tags:
1. May 31, 2017

### Mr. Chiappone

Say, we have two particles of equal and opposite charge in an isolated system in which we neglect gravity.

The energy of particle 1 is: E1 = U1 + T1

Where U1 is electrostatic potential energy and T1 is the kinetic energy of particle 1.

The energy of particle 2 is: E2 = U2 + T2

Where U2 is electrostatic potential energy and T2 is the kinetic energy of particle 2.

Therefore, the total energy of the system should be: E = E1 + E2 = U1 + U2 + T1 + T2

Where U = -kq^2/r

If we let the kinetic energy of both particles be zero then: E = U1 + U2 = -kq^2/r -kq^2/r = -2kq^2/r

The electrostatic potential energy of this system of two point charges is the energy (E = -2kq^2/r) when we neglect T (set T=0).

However, according to my book "Introduction to Electrodynamics" by David Griffiths, and Wikipedia "https://en.wikipedia.org/wiki/Electric_potential_energy", this is incorrect. Both of these sources would claim that the electrostatic potential energy of this system is -kq^2/r not 2kq^2/r. I am incredibly confused as to why they believe this or how my analysis could possibly be incorrect. It's not too hard to believe that Wikipedia is wrong, but my book for the two courses I took in electrodynamics is wrong too? That is hard to believe, and so I'm looking for another's perspective.

2. May 31, 2017

To get the potential energy of the system, you put one charge at the origin, and move the other in from infinity. You need to exert a force on the first particle to maintain its position, but no work is required in doing this. The potential energy is simply computed by the work that is done in moving the second charge. $\\$Alternatively, if you want to move both charges, one from $x=-\infty$, and the other from $x=+\infty$ with the final locations being $x=-r/2$ and $x=+r/2$, you will get two integrals for the work done. I'll just write one of them: $W1=\int\limits_{-\infty}^{-r/2} \frac{Q_1Q_2}{(2x)^2} \, dx$ , and note the distance they are separated as you bring them together is $|2x|$. If you evaluate the integral, and multiply by 2, you get the same answer as you do if you start with one charge at the origin. $\\$ Note: I left off the $k$ here and I am using cgs units.

Last edited: May 31, 2017
3. Jun 1, 2017

### Mr. Chiappone

Your reasoning makes sense, and I agree. So the next question is: why is my reasoning wrong?

Your conclusion would suggest that my definition of U1 and U2 and therefore E1, E2, and E are also incorrect.

Instead, we have (let particle 2 be a source charge that we fix in place, and let particle 1 be the test charge): E1 = T1 + U

And: E2 = T2

Thus: E = E1 + E2 = T1 + T2 + U

Would you agree?

If we were to instead bring both particles in from infinity as you did then:

E1 = T1 + U/2

And

E2 = T2 + U/2

Thus: E = T1 + T2 + U

Thank you for your input. I thought my reasoning was wrong, but my sources didn't explain why, they instead simply wrote down the answer.

4. Jun 1, 2017

### vanhees71

There is no potential except in static situations. Then the total energy of the system is (in Heaviside Lorentz units)
$$E_{\text{em}}=V(\vec{x}_1,\vec{x}_2)=\frac{q_1 q_2}{4 \pi |\vec{x}_1-\vec{x}_2|}.$$
In a certain non-relativistic approximation, i.e., for slowly moving charges, you can use the Lagrangian
$$L=\frac{m_1}{2} \dot{\vec{x}}_1^2 + \frac{m_2}{2} \dot{\vec{x}}_2^2-V(\vec{x}_1,\vec{x}_2).$$
The "self energies" are constant (and diverging for point particles) and thus not contributing in any way to the motion of the particles.

5. Jun 1, 2017

### Hardik Batra

[QUOTE="
The electrostatic potential energy of this system of two point charges is the energy (E = -2kq^2/r) when we neglect T (set T=0).
[/QUOTE]
Electrostatic potential energy means potential energy of charge particles which are at rest.
Initially suppose, both the charge particles are at infinite distance.
Suppose, You bring the charge q from infinity to let's say point A. No work has to do.
Now bring the another charge q from infinity to point B. Then you have to do work against the field of charge q at A.
And that much energy will store in the charge as Electrostatic potential energy.

Total energy U= U1 + U2 = 0+U1 = -kq2/r

6. Jun 1, 2017

### Mr. Chiappone

It's good to add additional information to the thread, thanks. Charlie told me what I needed to hear, but all of this is useful too for the other readers.

I agree with all of those statements.

7. Jun 2, 2017

### vanhees71

As you see from my posting #4, the potential energy is not of the form $V_1+V_2$. It's an interaction potential not the potential for charges in an external field!

8. Jun 19, 2017

### Mr. Chiappone

I suppose I should point out my deeper concern here that has lead me to question this topic, it's actually interesting, and has caused me some confusion.

Lets say we have a charged point particle in classical physics that emits radiation of some energy E, and momentum p, all in the positive x direction. Now, at the same time, this point particle absorbs radiation of energy E at the same instant that the said radiation was emitted, also, with all of its momentum in the positive x dimension. So, in this situation the charged particle has experienced no change in momentum.

Now, that said charged particle would remain stationary, would it not? I don't think that's much of a question considering the fact that the particle has experienced no change in momentum. Furthermore, had it not absorbed said radiation, this charged particle would have recoiled to conserve momentum, agreed?

Now, lets look at the situation we've just discussed. If I have the previously stated two particle system, and the test charge moves closer to the source charge, but the source charge remains stationary, then has the source charge absorbed radiation so that it may remain stationary? You see, I am concerned that if we are to force or state that one particle is to remain stationary as we move the other (increase or decrease the separation distance) then we need to transfer energy to the particle that we have chosen to fix in place. In which case, we are no longer talking about the energy of a two particle system, but the energy of a two particle system plus the energy that we transferred to the fixed particle so that it remains stationary.

This perspective is why I've developed doubts regarding Charles's approach, which is of course the standard conclusion regarding this topic.

From another point of view: In reality, if both particles are initially stationary, and in the next moment one is fixed while the other has gained some velocity, then the momentum of the system has not been conserved since in the first moment the total momentum was zero while in the next moment the total momentum is nonzero (since in the first moment, both particles are at rest, but in the next moment only one particle is moving). However, if we conclude that an outside entity has added momentum to the system such that the fixed particle remains stationary, then the momentum of the system would in fact be conserved since the system now includes the momentum of this outside entity (this is actually equivalent to the previous objection, but from another perspective).

In reality, if we were to observe the actual interaction, we would observe that both particles would accelerate simultaneously, and would have a velocity that is equal in magnitude, but opposite in direction. To keep one particle fixed we would need to either be in an accelerating frame of reference, or we would need to exert a force (and transfer energy) on the fixed particle.

To put this in another perspective, we should ask the question: What is potential energy?
Answer (statement 1): It is the potential of a particle to gain kinetic energy which exists whenever there is a nonzero force acting on the said particle.

When we state that we are going to fix a particle (source charge) in place while we move another (test charge), we are ignoring the fact that the source charge has a force acting on it. This is equivalent to setting the force on the source charge equal to zero which can only be achieved if we exert a force on the source charge such that it exactly cancels the force acting on it that's created by the test charge. In other words, we are removing the nonzero force and the potential energy of the source charge by simply stating that it is to remain fixed in place. If statement 1 is correct, then demanding that the force on the source charge is zero (demanding that the source charge remains fixed) is equivalent to removing the source charge's potential energy (because the force on the source charge is nonzero otherwise) via the intervention of an outside entity.

In terms of energy this looks like: E1 = T1 + U1 -> E1 = T1 + U1 + A1 = T1

Where A1 = -U1, and A1 is the energy we have added to the system (notice that A1 is positive as expected).

This lead me to consider the possibility that the energy of the system without the added energy is actually:

E = E1 + E2 = T1 + U1 + T2 + U2

This is the conclusion I currently come to when I ask the questions: What is the energy of particle 1? What is the energy of particle 2? If they have equivalent rest energy, and are initially at rest, then it is clear that they must have equivalent energy at all times. This must be true because they must have equivalent momentum at all times so that momentum is conserved (unless we intervene with an outside force).

But if we fix the source charge, it then becomes:

E = T1 + U1 + A1 + T2 + U2 = T1 + T2 + U2 = T1 + T2 + U

In other words: E1 = T1 + U1 -> E1 = T1

Do you see my point yet? I suppose I'll continue to elaborate for clarity.

If in reality, both particles must have equal energy at all times (otherwise momentum is not conserved) then U1 = U2, and T1 = T2.

So how do we calculate U1 and U2?

Well the force on both particles has a magnitude of: F = kq^2/r^2

Where r is the separation distance.

This implies that: U1 = U2 = -kq^2/r + constant

This becomes more clear if we think about this in terms of energy flow. Simply draw it out in your head. We know that to keep the source charge fixed in place, we must exert a force on the source charge via an outside entity. To make things simple in our minds, lets say that we use a counteracting electromagnetic force to do so. In this case, electromagnetic radiation is flowing to the source charge from an outside entity, and then from the source charge to the test charge all the while energy is flowing from the test charge to the source charge such that the source charge is neither gaining nor losing velocity. The point is... In order to keep the source charge fixed in place we must use a force from an outside entity (in this case the electromagnetic force), which means that energy is flowing to the source charge from that said outside entity in the form of electromagnetic radiation. Which means that we are not talking about the energy of a two particle system, but the energy of a two particle system plus the energy contributed via this outside entity.

I suspect the possibility that this reality is commonly overlooked due to the fact that the source charge, or the source of the gravitational field in some arbitrary problem, typically has a mass that is much greater than that of the test particle. This reality results with our ability to ignore the motion of the source of the force field without any serious consequences in our calculations because it is insignificant in comparison to the motion of the test particle. In other words, in all of the typical text book problems that we come across, fixing the source in place does not affect the motion of the test particle. For example: A satellite revolving around the Earth has a negligible effect on the motion of the Earth, and so we may consider the Earth to be fixed in place for our purposes.

Certainly, I'd like to simply accept Charles's approach and call it a day, but I have serious objections that do not seem to be resolvable. If you agree or disagree with this for whatever reason then be sure to let me know. I'm here for opinions after all, just be sure to back it up with a mathematical rebuttal that's relevant to the said claims.

9. Jun 20, 2017

### jbriggs444

To paraphrase, we have a charged point particle which has received a quantum of radiation from behind and emitted an identical quantum ahead. Equivalently, the charged particle is just sitting there while a quantum whizzes past. [Treating quanta as if they are merely little bullets -- not exactly accurate].

Yes, it is just sitting there while something has whizzed past.

The "source" charge is just sitting there, held in place by an unspecified mechanism. That mechanism requires no energy because it does no work. Any force that it exerts is applied over zero displacement.

No more than you need to transfer energy to a book sitting on a shelf in the library in order to fix it in place.

10. Jun 21, 2017

### Mr. Chiappone

Unfortunately, it seems you have seriously misunderstood me. First of all, I never mentioned quanta nor bullets so I don't know where you got that from. I mentioned radiation, that's correct, but we're talking about classical physics, not quantum mechanics. There's nothing wrong with considering the momentum of radiation.

It sounds like you missed the point. The point is that we have added energy to the system via an outside source to keep the particle fixed in place. Also, those were rhetorical questions, if you didn't notice...

So you think that the mechanism requires no energy to fix the particle in place? Although you had just agreed that the source charge will recoil without the intervention of this mechanism... You think this mechanism is capable of fixing an otherwise recoiling particle in place without exerting any energy onto the system? Think about that for a minute... I don't think you've thought that one through. This mechanism must exert a force on the recoiling particle to fix it in place. You cannot impart a force on a system without adding energy to that said system. If you want to think about quanta then think about what must carry the force... Force carrier particles right? Force carrier particles have energy. So if I impart any force on any system then I am also adding energy to that system because a force carrier particle carries energy.

A measure of work done on a particle is a measure of a change in potential energy. The claim is that the mechanism is adding energy to the system as it acts on the source charge, and this quantity of energy exactly cancels the potential energy of the source charge... Remember that the potential energy of the source charge is equal to the integral of the force that is acting on the particle over a distance from infinity to some point (call it r). So if the mechanism is acting on the source charge so that this force is equal to zero at all times then the mechanism is altering the potential energy of the source charge so that it is zero at all times (since the definite integral of zero is equal to zero).... So..... Of course the work done is equal to zero.... because the potential energy remains constant (and equal to zero) at all times. That reality obviously does not contradict the point I made.

Again, unfortunately, you're way off base. A book on a shelf has a total net force of zero acting on it. This said source charge without a mechanism to hold it in place has a nonzero force acting on it. So... a book and a source charge in a two particle system are entirely different on all levels.

Then again, the mechanism could be considered as the book shelf lol. What's the potential energy of the book without the shelf to hold it in place? It's the said integral of the force acting on the book. So what's the potential energy of the book that is placed on a shelf? It is again, that same integral, but this time the integral is equal to zero because the net force acting on the book is zero.

11. Jun 21, 2017

### Stephen Tashi

As a generality, the total energy of two combined physical systems is not the sum of their individual energies. Energy is not, in general, an additive quantity.

When their is no interaction between the two systems, it is safe to compute the total energy as the sum. For example, in thermodynamics the energy of two combined systems is often computed by addition because the energy of interaction between the two systems is small compared to their internal energies.

12. Jun 21, 2017

### jbriggs444

Indeed. Nothing wrong with considering bullets, packets of radiation, rocket motors or any similar thing.

Fair enough. We have a particle which is acting like a rocket. And we are supporting it with a spray of radiation (like a beach ball kept aloft with the spray from a garden hose).

You make the point that this support mechanism requires energy. Yes, it does.
No. I do not think that this mechanism is capable of doing so. I merely think that a mechanism is capable of doing so.

The notion of "exerting energy" is incoherent.

That is incorrect.
No. They do not have energy. In the heuristic description where we think of these as particles, they are virtual particles, not real particles.

If we hold the source charge in place, the integral is over a distance from 0 to 0. The integral is zero because the interval is zero, not because the force is zero.

The potential energy of the system consisting of the source charge and the test charge is the sum of the integral evaluated for the source charge and the corresponding integral evaluated for the test charge. The integral for the source charge is zero as I have just explained. The integral for the test charge is easily evaluated against a fixed coordinate system with a conservative force field -- that's the point of fixing the source charge in place.

A book on a shelf has two forces acting on it. Gravity and the support from the shelf. Neither does any work as long as the book stays in place.

A rocket-particle-beach-ball held in place by its emission of radiation in one direction and its receipt of radiation in another has two forces acting on it. The rocket motor acting in one direction and the incoming radiation in the other. Neither does any work on the rocket-particle-beach-ball as long as it stays in place.

That is not a correct understanding of potential energy. The potential energy of gravity considers only the force of gravity. It is the same with or without the shelf.

In any case, whether the shelf is there holding the book in place or absent, allowing the book to fall, the shelf does no work on the book.

Last edited: Jun 21, 2017
13. Jun 21, 2017

### Mr. Chiappone

Ok, then you agree that this mechanism is contributing energy to the system. That's the whole point because if that is true then claiming that a particle is fixed in place is a statement that subtly adds energy to the system via this mechanism that's acting on the fixed particle.

It's rare to debate such a thing. Virtual particles do in fact carry energy. The big difference between a virtual particle and real particle is that we don't observe a virtual particle. They're also able to momentarily violate conservation of energy over small intervals of time, but that's not too relevant to the topic... How could a virtual particle have any effect on a system if it did not carry energy or momentum? I'm not sure where you're getting this idea from, but you can explain if you'd like to.

No, the purpose of the mechanism is to make the force acting on the particle equal to zero, why else would we need it? The particle doesn't move because there's a net force of zero acting on it.

The sum of the evaluated integrals? You should be more specific. The integral you previously spoke of is an equation for work, correct? That equation is a measure of a change in potential energy. However, if we take one of the points in space to be r = infinity then we simply get an equation that's equal to the potential energy. However, the equation is then an equation that is equal to the work done on the particle from r = infinity to some point r... You should explicitly write out this math... you'll find an inconsistency in your reasoning.

Remember that the force, and the potential energy depends on distance alone. So it shouldn't be necessary to fix any particle in place because the motion of said particle does not actually come into an equation for potential energy, only the separation distance.

The book and the rocket-particle-beach-ball (lol) stay in place because there is a net zero force of zero acting on these objects in these scenarios. You're right though, the work done on these objects is equal to zero, but again, that does not contradict my point if you read it carefully enough.

The force acting on any object is equal to the gradient of the potential energy of said object... do you debate this? You can easily look it up in a google search.

Simply integrate that equation for force and you get potential energy... do you debate this?

If you agree with both of those statements then it should be clear that if the net force acting on an object is zero then the potential energy of said object is zero because you are integrating zero.

14. Jun 21, 2017

### jbriggs444

The fact that one mechanism wastes energy to hold an object in place does not mean that all mechanisms must do so.
You speak as if energy were a frame invariant thing. That is not correct.
Momentum and energy are separate things. Momentum added to a system can either add or subtract energy depending on one's choice of frame of reference.

Note that you claimed that your argument was intended to apply classically. Virtual particles do not exist classically.

If there is no net force acting, the particle does not move. If it does not move, the integral is zero. Pointing to the integral and saying that it is zero because the force is zero is irrelevant. Potential energy has nothing whatsoever to do with net force. It has to do with the force for which the potential is being evaluated.

Sauce for the goose. Can you tell us the definition of a potential?

Potential in a static field has to do with position in that field. It is computationally convenient to write down a static vector field, e.g. $$f(x) = -\frac{\vec{x}}{|\vec{x}|^3}$$ and compute that potential of that field. [Apply appropriate units and constant multiples for the field strength and charge magnitudes].

The work done separating two free particles from a test separation to the reference separation is equal to...
the work done moving one particle from a fixed second particle at the test separation to the reference separation is equal to...
the negative of the potential energy of the two particle system at the test separation.

15. Jun 21, 2017

### Mr. Chiappone

No need to consider wasted energy. I tried to give an example of a mechanism that can hold a recoiling particle in place, that is a stream of radiation. I'll try to explain it again, perhaps I wasn't clear enough in my elaboration... If a particle emits radiation it will recoil, but if that particle simultaneously absorbs radiation of the same energy and momentum at the moment that it emits said radiation then the particle will remain in place... Therefore, a stream of radiation that a particle is absorbing can act as a mechanism that is holding the particle in place as the particle interacts with another particle.

Notice that this mechanism (stream of radiation) is adding energy to the two particle system. It's not possible to name a mechanism that can hold a particle in place without contributing energy to the particle if that said particle has a force acting on it via another entity. That's the point I'm making.

Yes, momentum and energy are different. We don't need to consider quantum mechanics or virtual particles, I said...

But of course, we don't and probably shouldn't even consider anything that has to do with quantum mechanics as it only complicates things here.

Actually, I did give you the definition.

If a particle has an electromagnetic and a gravitational force acting on it so that the net force acting on the particle is zero then what is the potential energy of this said particle?

Let the potential energy due to the gravity be U1.

Likewise for the EM force, U2.

U = U1 + U2 = 0

I.e, the potential energy of the particle is zero, and if you write it out in terms of forces you'll find that we can combine U1 and U2 so that U is equal to the integral of the net force acting on the particle... which is zero.

The mechanism we speak of is doing exactly this, it's countering the force that's acting on the particle so that the net force is zero... and so the potential energy of the particle is zero.

16. Jun 21, 2017

### jbriggs444

And that point is incorrect. That is the point that I am making. A library book resting on a shelf is a worked example.

17. Jun 21, 2017

### Mr. Chiappone

I see what you're saying, but the normal force that the shelf exerts on the book contributes to the potential energy of the book. It is technically an electromagnetic potential energy. The total potential energy of the book is still zero although the gravitational potential energy is nonzero.

18. Jun 21, 2017

### jbriggs444

That is not correct. The normal force from the shelf does not contribute to the gravitational potential energy of the book.
Write down the equations for the two relevant conservative vector fields, please. You will find that the repulsion between book and shelf is a very short ranged force with a very small potential.

19. Jun 21, 2017

### Mr. Chiappone

I never said that it did.

It's the integral of the force that the shelf exerts on the book. The forces are equal in magnitude, but opposite in direction and so they will of course contribute potential energies of equal and opposite sign. Since the forces are equal, the potential energies cancel.

20. Jun 21, 2017

### jbriggs444

Calculate again. The gravitational force has greater range, so its integral is vastly greater.