Question

Discussion on the energy associated with the electric field

Answer 1

When working with capacitors one deduces by dividing the potential energy of the system by the volume it encloses that the energy density of the electric field is given by \[\upsilon=\frac{1}{2}\epsilon_0E^2\]
Then this equation is used over and over again for systems different than capacitors. Is there any way that through some arguments using the potential energy of an arbitrary system of discrete or continuous charges we may find the same result?

Answer 2

the generic potential - energy relations ship falls out of Coulumbs law. we get V = ∫ E•dA where E and A are field and area vectors.

and where field E = F /q = kq/r^2

you can get a lot of mileage out of just that, and maybe throwing in a bit of Gauss' Law.

Answer 3

I should have mentioned Newton also and ideas of how work/ energy and forces relate. W = Fx. As I am sure you know very well.

Answer 4

Found a solution

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!