Can electric and magnetic forces act on their sources?

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?
• 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. – Aaron Stevens Mar 4 '18 at 12:42
• @AaronStevens I added more information. I still believe that the book was talking about the field acting on its sources. Thank you. – A Slow Learner Mar 4 '18 at 22:22
• 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. – Aaron Stevens Mar 5 '18 at 18:48

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 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.