Electrical potential energy stored in vacumm for a single point charge?

Question

I have come to know the electrostatic potential energy in vacuum is given by $${\frac{1}{2}} \epsilon_0\int d^3x {E^2} $$ and this energy is due to the mutual electrostatic coulomb potential energy.

So only a single point charge in space also contributes to some electrostatic potential energy according to $${\frac{1}{2}} \epsilon_0 \int d^3x {E^2} $$ as due to a single point charge also we have an electric field at every point in space.

Now I am not getting it physically, why should there be any Coulomb energy in this case as there is no other charge to provide the electrostatic coulomb potential energy?

Answer 1

The field itself carries energy. This is, in fact, a vital point because it can be shown that, if the momentum and energy carried by the fields isn't accounted for, electromagnetism would blatantly violate Newton's third law (and this doesn't have anything to do with special relativity per se).

Answer 2

Now I am not getting it physically why should there be any Coulomb energy in this case as there is no another charge to provide the electrostatic coulomb potential energy?

You're right. For point charge the formula $$ \int \frac{1}{2}\epsilon_0 E^2 d^3\mathbf x $$ gives infinite value and is thus unusable - infinite energy would mean one cannot do calculations with it. Indeed, one can derive the above formula only for regular (finite density) distributions of charge, not for point charges. For set of point charges at rest, based on Coulomb experiments we define energy as

$$ W = \sum_{a}\sum_b' \frac{1}{4\pi\epsilon_0}\frac{q_a q_b}{|\mathbf r_a - \mathbf r_b|} $$ (primes means only those $b$ are used for which $b\neq a$). This formula can be transformed into $$ W = \int \sum_{a}\sum_{b}'\mathbf E_a \cdot \mathbf E_b \,d^3\mathbf x $$ where $\mathbf E_a$ is field due to point charge $a$, and so on. This formula can be generalized to case when charges are moving (analogous magnetic term is added) to obey both Maxwell's equations and the Lorentz formula for EM force. See the paper

J. Frenkel, Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534.

http://dx.doi.org/10.1007/BF01331692

If the charge is alone in large region, according to this formulae it has zero EM energy since the summation contains no term; its only energy is the rest energy $mc^2$ due to its mass.

Answer 3

One way to physically understand the energy of the electric field of an isolated charge is as follows:

Suppose we took some energy and converted it into an electron and a positron. The energy required is finite. (This can be inferred from the finite energy produced by an electron/positron annihilation or from experiments that produce such pairs).

We can analyze this system using only electrostatics assuming

1. The electron and positron exist individually ie there is some spacing between them
2. The spacing is such that the laws of electrostatics are valid and all other forces are negligible
3. The electron and positron are at rest.

The field has some finite initial electric energy. We now hold the electron and move the positron to some far distance like infinity. The work so done adds to the energy of the electric field. Since the charges are far apart we can say that they are isolated and choose to divide the field energy between the electron and the positron.

So, the "isolated" charge has a field and the field has energy.

Answer 4

The field energy results from self interaction of the particle. It is usually considered to contribute to its mass and absorbed into it.

The problem is that for a point particle it is infinite and this is one of the infinities of QED. We can get away with it but renormalization is not fully satisfactory - in my opinion.