# Physical meaning of the energy density of an electrostatic field

I understand the physical meaning of electrostatic energy of a system of charges (or a distribution with given density) as the energy stored in the system while working to carry the charges from infinity to their actual place in the system. According to this article on Wikipedia, in the case of a static field you can also compute that energy as the integral of energy density $U=\int\frac{1}{2}\epsilon_{0}|\vec{E}|^2dV$. What is the physical interpretation of this density? What is the physical meaning of the expression energy of an electrostatic field and can this concept be used also in the non-static case? And with other fields as the gravitational one?

P.S. I hope this question doesn't seem obvious or useless. Being a student of mathematics, I really like to think about an abstract field $\vec{E}$ governed by Maxwell equations and then give it some physical meaning, unfortunatly I have not seen any theoretical physics yet, only some general physics.

-
On the contrary, I think this is a great question. This is the sort of thing physicists don't often think about. – David Z Jan 8 '12 at 0:30

Actually in electrostatics energy density of E-field is not a physical observable. As you say, only when charges move will there be any work done. While the two ways of calculating total energy end the same, you cannot distinguish whether energy is stored on the charges or in the field. Even E-field itself is more of an abstract mathematical entity, without which everything can be calculated in terms of Coulomb law.

The physical reality of E and B fields (and the energy density associated) becomes apparent only in non-static cases. For example, in electromagnetic radiation, fields can propagate in free space without associating with charges and currents, and the radiation may do work on non-charges (for example, light pressure). Because from Maxwell equations we can derive a general formula of energy density

$$\rho = \frac{\epsilon_0}{2} |\vec E|^2 + \frac{1}{2\mu_0} |\vec B|^2$$

which coincides with the electrostatic case, we deduce that even in electrostatics energy is indeed stored in the fields.

-
Thank you, in the first part of the lecture course I followed we mostly covered static fields. Now I see the point – user6452 Jan 7 '12 at 22:59

Yes, $\epsilon \vec E \cdot \vec E$ is the electrostatic part of the energy density carried by the field. The energy density of the electromagnetic field also includes the magnetic term: $$\rho_{E,B} = \frac{\epsilon}{2} |\vec E|^2 + \frac{1}{2\mu} |\vec B|^2$$ and this formula is valid even for arbitrary time-dependent, variable electromagnetic fields. When you mentioned the energy density $$\frac 12 \int \rho_{\rm charge} \Phi \,\,dV,$$ one should note that one must be careful to avoid double-counting. When we assume that the energy is carried by the electromagnetic field, we should no longer add the $\rho_Q\cdot \Phi$ term separately because we could be double-counting. However, in some respects, they have to be separated and both of them have to be added.

At any rate, $\epsilon|E|^2$ is a term in the formula for the total energy, anyway. It's important to know because only the total energy, with all the terms that should be there, is conserved.

One may interpret the energy $\int dV\,\,\epsilon|E|^2/2$ as work, in the same way as for the interaction energy of the charges you mention. It's the work needed to change the electrostatic field from the situation $\vec E=0$ to the given configuration of $\vec E$. The energy may be given as an integral of the work, $$E_{\rm energy} = \int dV \int dt\, \vec E\cdot \frac{d\vec D}{dt},\qquad \vec D \equiv \epsilon \vec E$$ Note that there is no $1/2$ in the formula above; it comes from the integration. So the larger the field is at a given point, the harder it is to increase its value there.

-
I can't understand the last formula for the energy as an integral of the work. Furthermore, I can't understand what you intend for "work to change the field". The only way I can think about doing work is by moving charges. – user6452 Jan 7 '12 at 14:00
Dear Marco, "moving charges" is a mechanical work. But mechanical work is not the only type of work. Just like there's energy in electric field, there's el. work. A transformer is made out of 2 coils inside each other; one of them works to increase the magnetic field in the other which ultimately induces the current in the other coil; and although they're not mechanically connected (and they're not connected by conductors, either), it's possible to transfer energy between them. This energy, coming from the otherwise disconnected 2nd coil, may be later used to lift an elevator or anything else. – Luboš Motl Jan 7 '12 at 20:20
Quite generally, you are imagining that the energy and work have to be mechanical and the fields are "completely different". But they're not completely different. The energy density $E^2/2$ may be imagined as having a spring, energy like $kx^2/2$, at each point of space (or in a dense lattice): you just call it $E$ instead of $x$, and there are 3 springs per point, $E_x,E_y,E_z$. Then the changing of the energy in the electric field is the same kind of work as stretching a spring. These are words that may be disputed; the formulae express exactly what I mean and what is true. – Luboš Motl Jan 7 '12 at 20:28
I know that an electrical field can do some work $W=q\int \vec{E} d\vec{s}$ What I meant is: I can't figure out how an electrical field can do any work without moving charges. If I understand your example, charges are moving in the current of the coils. Should I interpret the expression "energy of the field" as: "ok, if there were some charges in this field, it would be capable to move them and use its energy to do that work"? – user6452 Jan 7 '12 at 20:44
Dear Marco, whether charges are ultimately moving while work is being done is completely irrelevant for the fact that the electromagnetic fields contribute to energy. The situation is fully symmetrical. You could have also said: "I can't figure out how charges may ever accelerate each other without changing the electromagnetic fields in between." It's just your irrational prejudice to think that the mechanical energy is the only right one or the primary one, to try to reduce everything to charges and mechanics. Treat the formulae mathematically! The electric and mechanical terms are on par. – Luboš Motl Jan 8 '12 at 7:15

When one has a distribution of charges $q_1,\dots,q_n$ at points $\vec{r}_1,\dots,\vec{r}_n$, the energy of the system is given by the sum of the energy of each particle due to its interaction with the others divided by two since each interaction is counted twice i.e. $$U=\sum_{i=1}^n\sum_{\substack{j=1 \\ j \neq i }}^{n}\frac{1}{4\pi\epsilon_0}\frac{q_iq_j}{\|\vec{r}_i-\vec{r}_j\|}=\frac{1}{2}\sum_{i=1}^nq_i \phi_i(\vec{r}_i)$$where $\phi_i(\vec{r}_i)$ is the potential at $\vec{r}_i$ due to all charges except $q_i$. If we go over to a continuous charge distribution with charge density $\rho(\vec{r})$, the summation is replaced by an integration over "infinitesimal chunks of charge" $dq=\rho(\vec{r}) \textrm{d}V$. Then the energy of the system is $$U=\frac{1}{2}\int\limits_V \rho(\vec{r})\phi(\vec{r})\textrm{d}V$$Now, due to Gauss's Law for electricity, $\vec{\nabla}\cdot\vec{E}=\frac{\rho}{\epsilon_0}$ we have $$U=\frac{1}{2}\int\limits_V \epsilon_0\left(\vec{\nabla}\cdot\vec{E}(\vec{r})\right)\phi(\vec{r})\textrm{d}V$$Recalling the vector identity $\vec{\nabla}\cdot\left(f\vec{F}\right)=f(\vec{\nabla}\cdot\vec{F})+\vec{F}\cdot(\vec{\nabla}f)$ we have $$U=\frac{1}{2}\epsilon_0\left[\int\limits_V \vec{\nabla}\cdot\left(\vec{E}(\vec{r})\phi(\vec{r})\right)\textrm{d}V-\int\limits_V \vec{E}(\vec{r})\cdot\vec{\nabla}\phi(\vec{r})\textrm{d}V\right]$$By replacing the first integral over the volume $V$ for one over its boundary $\partial V$ through the divergence theorem which states $\int_V\vec{\nabla}\cdot\vec{F}\textrm{d}V=\oint_{\partial V}\vec{F}\cdot\textrm{d}\vec{S}$ and by remembering the definition of potential $-\vec{\nabla}\phi=\vec{E}$ we get $$U=\frac{1}{2}\epsilon_0\left[\oint\limits_{\partial V} \vec{E}(\vec{r})\phi(\vec{r})\textrm{d}\vec{S}+\int\limits_V \vec{E}(\vec{r})\cdot\vec{E}(\vec{r})\textrm{d}V\right]$$ Now we may choose the volume of integration $V$ to be all space. Then $\partial V$ would be infinitely far away from all charges and by convention the potential $\phi$ would die out, making the first integral to dissapear. Then we would be left with the following expression for the energy of the system $$U=\frac{1}{2}\epsilon_0\int\limits_V \|\vec{E}(\vec{r})\|^2\textrm{d}V$$ From this expression follows that the energy density $\Upsilon$ at $\vec{r}$ is given by $$\Upsilon(\vec{r})=\frac{1}{2}\epsilon_0\|\vec{E}(\vec{r})\|^2$$Now we may ask "where" is this energy. Note we derived it as the energy density of the total energy of the charges in our universe. Non the less, this energy density seems to be distributed even where there may be no charges. So we ask the question, does the energy belong to the charge configuration or to the electric field? From the point of view of electrostatics, both are equivalent, and in our last equation, we are inclined to think of the electric field as being the carrier of energy. In electromagnetism, this is a necessity, since the electric field may exist and propagate quite independently of its source charges in the form of electromagnetic waves, for example light, which evidently carries energy, since most of the energy in planet earth is carried from the sun in this form. Any other question, please ask! I'll try to give an answer if I know one!

-
Very good answer. Can the same proof be adapted for a discrete, linear or spatial distribution of charge? I've never found a proof for such distributions of charge... Thank you so much!!! – Self-teachingDavide Oct 6 '15 at 6:00
This proof actually works for a discrete set of charges. A charge density which describes a discrete set of charges such as the one described at the beginning of my answer is $\rho(\vec{r})=\sum_{i=1}^n q_i\delta(\vec{r}-\vec{r}_i)$. The problem with attempting to avoid the delta functions is that we don't have a better description of what the infinity of the coulomb forces looks like in the singularity of a charge. Hope that helps! – Iván Mauricio Burbano Oct 6 '15 at 10:03
Thank you for the answer! I realise that it's a case I cannot understand for now, with my limited knowledge of real analysis, since I know no proof of the mathematical facts you uses (like the divergence theorem) in the case of distributions such as the one defined by the use of the $\delta$... – Self-teachingDavide Oct 7 '15 at 8:12