Charges in free space

We have shown that given a charge distribution $\rho(\vec r)$, we have an electric field $\vec E(\vec r)$ and electric potential $\phi(\vec R)$ all over the space. The electric field $\vec E(\vec r)$ gives the force acting on a charge q situated at point $\vec r$,

$$\vec F = q \vec E.$$

If we move the charge from point $\vec r_1$ to $\vec r_2$, the work done by the force exerted by the charge distribution on charge $q$ is given by

$$W = \int_{\vec r_2}^{\vec r_1} d \vec l \cdot \vec F(\vec r)$$ (143)

where $\vec dl$ is an incremental line element along a path joining points $\vec r_1$ and$\vec r_2$. Then

$$= q \int_{\vec r_2}^{\vec r_1} \vec dl \cdot \vec E(\vec r)$$

$$= - q \int_{\vec r_2}^{\vec r_1} \vec dl \cdot \vec \nabla \phi(\vec r)$$

$$= q ( \phi(\vec r_1) - \phi(\vec r_2))$$ (144)

Thus, the work done on charge $q$ is $q$ times the potential difference between two point$\vec r_1$ and $\vec r_2$. Note that the work done does not depend on the details of the line joining the two points. This also means that $q \phi(\vec r)$ gives the energy required to bring the charge $q$ from infinity to $\vec r$. We can therefore compute the energy required to assemble the charge distribution $\rho(\vec r)$.

Consider discrete charges first. The energy required to bring $q_i$ to a point $\vec r_i$ is

$$E_i(\vec r_i) = q_i \sum_{j \ne i} \frac{q_j}{\vert \vec r_i - \vec r_j\vert}.$$

Then the total energy to assemble all the charges is

$$E = \frac{1}{2} \sum_{i,j, i \ne j} \frac{q_i q_j}{\vert \vec r_i - \vec r_j\vert}$$ (145)

The factor of half appears because each of the charge is counted twice in the sum above. Also note that $i=j$ terms are not included in the sum. That is because this term essentially gives the energy required to assemble any point charge $q_i$, that is the self energy and that is infinite.

For continuous charge distribution $\rho(\vec r)$ the corresponding energy associated with it is

$$E = \frac{1}{2} \int d^3r_1 d^3r_2 \frac{\rho(\vec r_1) \rho(\vec r_2)}{\vert\vec r_1 - \vec r_2\vert}$$

$$= \frac{1}{2} \int d^3r \rho(\vec r) \phi(\vec r)$$

$$= - \frac{1}{8 \pi} \int d^3r ( \nabla ^2 \phi(\vec r) ) \phi(\vec r)$$

$$= - \frac{1}{8 \pi} \int d^3r \phi (\vec r) \vec \nabla \cdot \vec E(\vec r)$$ (146)

The first line in the equation above is just the generalization for continuous charge distribution. The second line follows from the definition of the potential in terms of the charge distribution and finally the last line is the result of Poisson equation. Note that here the self energy is not excluded. Also note that the energy is now expressed in terms of the potential and there is no reference to the charge distribution. One can perform a partial integration and express the energy in terms of the electric field alone. The result is

$$E = \frac{1}{8 \pi} \int d^3r ( \vec \nabla \phi(\vec r) )^2$$

$$= \frac{1}{8 \pi} \int d^3r E^2(\vec r)$$ (147)

The partial integration has a surface term ( $\int_V d^r f(\vec r) \vec \nabla \cdot \vec g(\vec r) = \int dS f(\vec r) \hat n \cdot g(\vec r) - \int d^3r ( \vec \nabla f(\vec r)) \cdot \vec g(\vec r)$ ). For any confined charge distribution, the potential at infinity goes at least as $1/r$ and therefore the surface integral vanishes for the surface at infinity.

This is a very important result for several reasons. First note that the energy is expressed in terms of the electric field and the charges do not appear in the equations. So, the electric field has meaning of its own. In Newtonian mechanics, a force is exerted on a body by external objects and there is no reference to the field. That is there is the idea of action at a distance. With the definition of electric field there is a conceptual change. A charge produces an electric field around it and this field interacts with another charge. So there is no action at a distance. The field interacts with the charge locally. This has very important consequence from the point of special relativity. Later we shall see that the concept of simultaneity is not valid in special relativity and therefore the concept of action at a distance, which implies simultaneity in all inertial frames, is not consistant with special theory of relativity. On the other hand, the concept of field has no difficulty with relativity because the field not instantaneously generated by a charge and the interaction of field with a charge is local, so there is no problem with simultateity.

Another point is, one can think of electric fields without any reference to charges So electric field has an existance of its own. Later we shall find that we can indeed have electric fields without any ( apparant ) charges as in case of electromagnetic waves. And the part of the energy of the electromagnetic waves is due to the electric field in it and that energy is exactly what we have obtained above.

Third thing to note is that the energy in above equation is positive definite. That seems to be counter-intuitive because if we bring a positive and a negative charge distributions together, energy would be neative because opposite charges attract. The reason for this difference is that the above expression includes the energy due to self interaction, i.e. the energy required to assemble the individual charge distributions. This term was not included in the computation of energy of point charges because there the self energy is infinite. For continuous charge distributions this is not so.