# Misconception about electrostatic energy

The electrostatic energy of a charge distribution $\rho$ is

$$U=\int\rho(\vec{r})\ \phi(\vec{r}) \ dV$$

where $\phi$ is the electric potential generated by the charge distribution.

Also, after doing some math, one can arrive to this other expression

$$U=\frac{1}{8\pi}\int|\vec{E}|^2 \ dV$$

where $\vec{E}$ is the electric field generated by the charge distribution.

This two expressions are said to be just two ways of calculating the same thing: the energy of the charge distribution. However, it is also said that the second formula means that the electric field can store energy. So, what is $U$? is it the energy of the charges? Is it the energy of the electric field the charges generate? can an electric field store energy if there are no sources generating it? (for example: In radiation).

I don't understand how much Independence the electric field has with respect to the charges generating it. Is $\vec{E}$ a new entity with its own energy or is U relating $\vec{E}$ and $\rho$ in some way, like an interaction energy between the electric field and the charged particles?

can an electric field stock energy if there are no sources generating it? (for example: In radiation).

The two equations here are equivalent, and so they have the same implications and meaning. Since the energy obviously vanishes in the first equation in the absence of sources (i.e. when $\rho=0$), it also vanishes in the second equation.

This two expressions are said to be just two ways of calculating the same thing: the energy of the charge distribution. However, it is also said that the second formula means that the electric field can stock energy. So, what is UU? is it the energy of the charges? is it the energy of the electric field the charges generate?

You just said what this is: it's the energy of the charge distribution. The energy is generated through electrostatic interaction, so, sure, it's also the energy of the electric field. Those are the same thing.

I don't understand how much independance does the electric field has with respect to the charges generating it. Is $\vec{E}$ a new entity with its own energy or is $U$ relating $\vec{E}$ and $\rho$ in some way, like an interaction energy between the electric field and the charged particles?

The derivation of the second formula requires the fact that $$\textrm{Div}\left(\vec{E}\right) = \frac{\rho}{\epsilon_0}.$$ The electric field is entirely determined by the charge distribution.

• Addendum: Actually, radiation does store energy without a need for a non-zero charge distribution, but this involves electrodynamics (rather than electrostatics). In the non-static case the second formula in the original question is extended by a $|\vec B|^2$ term and therefore the first formula should also look differently and give a non-zero result even if the charge density is zero. – Photon Jul 25 '17 at 20:22

the energy u is the entire Volume covered by the integration . The field's strength E is a force but when squared it gives the energy density(The energy per unit volume of space) but not the total energy.

The two total energy equations can be shown to be numerically equal by integration by parts. This is demonstrated in Feynman volume II-8-11.