Physics with Calculus/Electromagnetism/Field Energy

Taken from: https://en.wikipedia.org/wiki/Electrostatics#Electrostatic_energy

A single test particle's potential energy, $U_{\mathrm{E}}^{\text{single}}$, can be calculated from a line integral of the work, $q_n\overrightarrow{E}\cdot \mathrm{d}\overrightarrow{\ell }$. We integrate from a point at infinity, and assume a collection of $N$ particles of charge $Q_n$, are already situated at the points ${\overrightarrow{r}}_i$. This potential energy (in Joules) is:

$U_{\mathrm{E}}^{\text{single}}=q\varphi (\overrightarrow{r})=\frac{q}{4\pi {\epsilon }_0}\sum _{i=1}^N\frac{Q_i}{\Vert {\overrightarrow{𝔯}}_i\Vert }$

where ${\overrightarrow{𝔯}}_i=\overrightarrow{r}-{\overrightarrow{r}}_i,$ is the distance of each charge $Q_i$ from the test charge $q$, which situated at the point $\overrightarrow{r}$, and $\varphi (\overrightarrow{r})$ is the electric potential that would be at $\overrightarrow{r}$ if the test charge were not present. If only two charges are present, the potential energy is $k_e{Q}_1{Q}_2/r$. The total electric potential energy due a collection of N charges is calculating by assembling these particles one at a time:

$U_{\mathrm{E}}^{\text{total}}=\frac{1}{4\pi {\epsilon }_0}\sum _{j=1}^N{Q}_j\sum _{i=1}^{j-1}\frac{Q_i}{r_{ij}}=\frac{1}{2}\sum _{i=1}^N{Q}_i{\varphi }_i,$

where the following sum from, j = 1 to N, excludes i = j:

${\varphi }_i=\frac{1}{4\pi {\epsilon }_0}\sum _{j=1(j\ne i)}^N\frac{Q_j}{4\pi {\epsilon }_0{r}_{ij}}.$

This electric potential, ${\varphi }_i$ is what would be measured at ${\overrightarrow{r}}_i$ if the charge $Q_i$ were missing. This formula obviously excludes the (infinite) energy that would be required to assemble each point charge from a disperse cloud of charge. The sum over charges can be converted into an integral over charge density using the prescription $\sum (\cdots )\to \int (\cdots )\rho {\mathrm{d}}^{3}r$:

$U_{\mathrm{E}}^{\text{total}}=\frac{1}{2}\int \rho (\overrightarrow{r})\varphi (\overrightarrow{r}){\mathrm{d}}^{3}r=\frac{{\epsilon }_0}{2}\int {\left|𝐄\right|}^2{\mathrm{d}}^{3}r$,

This second expression for electrostatic energy uses the fact that the electric field is the negative gradient of the electric potential, as well as vector calculus identities in a way that resembles integration by parts. These two integrals for electric field energy seem to indicate two mutually exclusive formulas for electrostatic energy density, namely $\frac{1}{2}\rho \varphi$ and $\frac{{\epsilon }_0}{2}{E}^2$; they yield equal values for the total electrostatic energy only if both are integrated over all space.