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Consider two stationary charges, one positive the other negative. Their potential energy is clearly negative. So you would expect that the energy density of the associated electric field would also be negative. But it isn’t. It’s the square of the electric field and therefore positive. Why isn’t the energy density negative like the potential energy?

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""Consider two stationary charges, one positive the other negative. Their potential energy is clearly negative. "" Really? –  Georg Oct 17 '11 at 10:29
    
U(r)=kQq/r < 0, if signs of Q and q differ. –  Belizean Oct 17 '11 at 10:41
    
Belizean means interaction energy of two charges $U(\vec{r}_1-\vec{r}_2)$ which is, of course, can be negative. –  Vladimir Kalitvianski Oct 17 '11 at 21:11
    
Vladimir and Jing: Your point is that the field energy is distinct from the interaction energy. That seems to make sense, if you consider the approach of charges q and -q. Their interaction energy becomes arbitrarily negative, while their field energy goes to zero. But isn't the interaction energy just the work required to bring the charges in from infinity to their particular (stationary) separation? And isn't it straight forward to show that this work is in fact equal to the field energy? –  Belizean Oct 18 '11 at 16:17
    
@Belizean: OK, let us take a single charge. It creates an electric fields which has energy density and the total energy. Let us suppose the charge is not point-like and the field energy is not infinite. What to do with this energy? Nothing. It comes along whenever the charge goes. We do not use it. It only appears in funny self-action ansatz and is discarded in the end (mass renormalization). On the other hand, the propagating field energy is more interesting - it does not stay attached to the charge. –  Vladimir Kalitvianski Oct 18 '11 at 22:56

4 Answers 4

up vote 2 down vote accepted

The electric field energy is indeed positive for the case of two opposite charges, but it is smaller than the electric field energy when the charges are separated. The difference is the potential energy of interaction of the two charges, which includes the self-energy, the field energy for a single charge, inside the mass of the charge.

You can see that the energy is decreasing when you bring to equal charges together, because when they are right on top of each other, the field is zero. To see mathematically that the decrease is the potential energy, consider the energy integral

$$ \int |\nabla \phi|^2 dV = - \int \phi \nabla^2 \phi = \int \phi \rho $$

Where the first equality is an integration by parts, and $\rho$ is the charge density. When you can write the field $\phi$ as a sum of two contributions from two separate small charged spheres, the pontential energy of the mutual interaction is given by the charge of one times the potential generated by the other. There is also the self-pontential, which is the energy of the charge in its own field, which is classically divergent for a point charge, and can be neglected because you assume the charge radius is unchanged as the particles move around, so that this energy is constant.

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The self-energy contribution to your integral makes it positive. Discarding it is equivalent to discarding the notion of field and leaving only action at a distance between charges. –  Vladimir Kalitvianski Oct 19 '11 at 14:11
    
@Vladimir: yes, exactly--- this is the whole point of the interaction potential energy. I hope it is clear enough in the answer as it is. –  Ron Maimon Oct 19 '11 at 17:02
    
Ron Maimon: Your answer is precisely what I was looking for. The answer, then, is that the field energy is positive despite the interaction energy ("potential energy") being negative, because the field energy includes the self energy of the charges. Also, you made a sign error in your answer. [Poisson's equation for the electric potential is sourced by negative charge density.] [I would liked to have commented in the appropriate place, but I seem to have lost that ability and am too stupid to figure out how to restore it.] –  Belizean Oct 19 '11 at 19:36
    
It's not your fault--- you don't have enough reputation yet to comment. I just fixed the sign. I also missed factors of 2 and pis which are conventional, I wrote it quickly to give the main idea. If you like the answer, you can click "accept" on it and that's that, you don't need to place a separate answer. Once you do this, you can delete this answer. –  Ron Maimon Oct 19 '11 at 19:56
    
yes, this happens often if you sign in twice. I'll flag the question for administrator attention. –  Ron Maimon Oct 20 '11 at 5:17

The field generated by the two charges $q_1$ and $q_2$ is the sum of the field generated by the first and the second charge, let's say $ \vec E_1+\vec E_2$. When you square this sum and integrate in the whole space you got three terms. Integrating all over the space the square of $ \|\vec E_1\|^2$ and $\|\vec E_2\|^2$ we obtain the self energy of the two charges on their own i.e. the energy of the field of a lone charge. Each of them is positive and the sum of these two is bigger than the integral of the double scalar product of $ 2(\vec E_1\cdot\vec E_2)$ in the whole space. This last is negative and is the usual energy obtained multiplying one charge for the potential generated by the other.

Nevertheless a question troubles me: why this is not in standard textbooks? Furthermore, not having a sound background in Physics, has somebody any idea of what is the physical meaning of this $ \|\vec E\|^2$ self energy. How does it relates with the mass of the electron and how this energy relates with the photons associated with the electromagnetic wave there should be every time a positron-electron couple rise.

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Hi @Newbie, and welcome to Physics.SE. Your second paragraph is a question unto itself, rather than an answer. It would be better (and more likely to get an answer) if you were to post it as a question. If you do, you should also probably edit this answer to remove that paragraph. –  Colin McFaul Jan 16 '13 at 20:11

Potential energy of interaction is about interaction and can be negative.

The electric field created by two charges for a third one is a vector sum of two fields:

$\vec{E}=\vec{E}_1(\vec{r}_3-\vec{r}_1)+\vec{E}_2(\vec{r}_3-\vec{r}_2)$.

It defines the corresponding force. From it you can derive the potential energy of interaction of the third charge with two original ones. Squared, it does not represent anything about interaction of the first two.

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Here, you should refer to the definition of energy density, the electric field energy density is defined as following:

$u = \frac{1}{2}\epsilon |\vec{E}|^2$

where $\epsilon$ is the permittivity of media.

therefore this energy density should always be positive.

The derivation of the electric energy density can be referred to the wiki post

btw, I think you made some mistake statement in your question on your example

"Their potential energy is clearly negative. "

We can look at you example of two charge particles, we can easily write out the potential everywhere in your case as

$\Phi(\vec{x}) = \frac{Q_1}{4\pi\epsilon |\vec{x}-\vec{r_1}|}+\frac{Q_2}{4\pi\epsilon |\vec{x}-\vec{r_2}|}$

where $\vec{r_1}$ and $\vec{r_2}$ are the position of these charges in space, if we assume $Q_1$ is positive, $Q_2$ is negative, then we can calculate the electric potential everywhere. Clearly, you'll notice that there is zero-potential surface perpendicular to the line connects these two charges, on the side of $Q_1$, potential is always positive, and on the side of $Q_2$, potential is always negative, then the electric potential energy at these two charges are both positive ($U_1 = \Phi_1Q_1 >0$, since $\Phi_1>0$, similarly, $U_2 = \Phi_2Q_2 >0$, cause $\Phi_2 <0$ and $Q_2<0$). So the electric potential energy is positive. Also you should note that this energy is only a part of total electric energy.

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