The following is quoted from Griffith's electrodynamics book, 3rd edition:

Suppose we have some charge and current configuration which, at time t, produces fields $\mathbf E$ and $\mathbf B$. In the next instant, $dt$, the charges move around a bit... According to the Lorentz force law, the work done on a charge $q$ is: \begin{split} \mathbf F \cdot dl &=q(\mathbf E + \mathbf v \times \mathbf B) \cdot \mathbf v dt\\ &=q\mathbf E \cdot \mathbf v dt\\ &=\rho dv \mathbf{E} \cdot \mathbf{v}dt\\ &=\mathbf{E} \cdot\mathbf{J} dvdt\\ &=\left[ \frac{1}{\mu}_0 \mathbf{E} \cdot (\nabla \times \mathbf{B})-\epsilon_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} \right ] dvdt \end{split}

The last equality comes from Ampere-Maxwell's law: the $\mathbf{J}$ in this equation must be the $\mathbf{J}$ of the source charges, which in turn gives rise to the source magnetic field $\mathbf{B}$, rather than the $\mathbf{J}$ of the external (test) charges under the influence of $\mathbf{B}$, since that is the meaning of the Ampere-Maxwell's law.

The above paragraph seems to suggest that the fields created by a particular charge and current distribution can actually exert forces on their sources. But as far as my understanding goes, a field only exerts forces on external charges (test charges).

My question is:

  1. Is there something that I am missing in interpreting the above paragraph?
  2. Can electric and magnetic forces act on their sources?
  • $\begingroup$ The whole idea of self inductance is the magnetic field produced by an object acting on the object. Is this what you mean? Plus the quoted paragraph isn't talking about a field acting on its source necessarily. You can have some configuration, and then have another charge in this configuration and look at the work done on it. $\endgroup$ Mar 4, 2018 at 12:42
  • $\begingroup$ @AaronStevens I added more information. I still believe that the book was talking about the field acting on its sources. Thank you. $\endgroup$ Mar 4, 2018 at 22:22
  • $\begingroup$ I still don't think this is talking about self-interaction. It is just talking about the fields at some point in time due to everything else. It is like the simple example of the electric force between two charged particles. Both particles contribute to the overall field in space, but to determine the force on one of the particles you just consider the field produced by the other particle. $\endgroup$ Mar 5, 2018 at 18:48

2 Answers 2


The section of Griffiths that you quote from says nothing about source or test charges. He is not distinguishing those at all. He is simply assuming charge and current are regular distributions so the expression $\mathbf E\cdot\mathbf J$ is integrable. In the discussion and the resulting theorem, the EM field and the sources are total - if you distinguish field of sources and field of test charges, he uses total field $\mathbf E = \mathbf E_{sources} + \mathbf{E}_{test}$.

If the question is whether, independent of this derivation, charges can act on themselves via EM forces, this is actually a subtle question that has no single answer. It depends on how we choose to model the charge distribution in space. More precisely, it depends on whether the electric field is finite everywhere in the region of the body. This in turn depends on how the charge is distributed in space.

If the charge is concentrated at a point or on a line, the electric field is not defined in those places at all. It is not possible to calculate electromagnetic self-force on a body that has this kind of geometry. One can still calculate the external force due to fields of other bodies, but one has to use external field, not total field (which is not defined). For these kinds of charge/current distribution, the Poynting theorem is valid only outside the distribution, so it cannot be used to relate work done on the charged body with EM energy. Mathematically $\mathbf E\cdot\mathbf J$ is not integrable in the place where the charges are.

If the charge is distributed on a plane or on a sphere, or throughout a 3D body, the electric field due to its charge is finite in the body, so force due to its EM field (so called self-force) can be calculated. For a sphere at rest or in rectilinear motion, this force is zero; but if it is accelerating, it is non-zero.

This self-force is what is damping charge motion in an antenna when it is radiating EM waves - the self-force is an explanation for radiation damping. One can view it as due to action of all (non-point) parts of an antenna on all other (non-point) parts.

Another example of self-force is electromagnetic self-induction effects. When current changes in a coil, a vortex-like electric field due to those charges is present in the nearby space. This field acts back on the charges forming the current to resist the original change. This results in a kind of inertia of electric current.


If you consider a continuous distribution of charges and currents then the answer is definitely YES

Electric and magnetic forces do act on their sources.

Any infinitesimal part of the distribution feels the fields created by the entire distribution, because its own contribution part in this field is infinitesimal and the question whether it feels its own contribution or not is moot, just because its own contribution is infinitesimal !

The difficult question is about "point charges". The field created by a point charge is infinite at the position of the point charge itself. However, it does not have a specific direction ! Near the particle, the field is always in the direction of the point charge, and goes to infinity in different direction when you get close to the point charge itself coming from different directions. So inasmuch as the field of a point charge to itself is infinite but with no defined direction it is omitted, essentially because it does not make sense.

I can understand you don't feel happy with this answer, but it is the only one that makes sense.

In (relativistic) Quantum Mechanical field theory a more precise explanation can be presented. It is the very complicated problem of the so-called "charge renormalisation", but this is way above what can be explained here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.