Does a charge get accelerated by the field it creates? and understanding the derivation of Maxwell's stress tensor The total electromagnetic force on a charge in volume is given by:
$$F=\int_V (E + v\times B)\rho \ d\tau$$
and thus the force per unit volume is:$$f=\rho E + J\times B$$
At this step, since we want to write this in terms of fields alone, we write $$\rho=\epsilon\nabla\cdot E$$ and substitute (and similarly for $J)$
What is not clear to me is that how come the $E$ in the first equation same as the $E$ created by the charge density it acts on? Shouldn't there be $E_1$ and $E_2$ ?
My guess is that the reason we can do this is because we are talking in terms of charge distributions which is to say that the field $E$ acted at a point by the surrounding charge is not different from the field acting at any other point by the charge surrounding it but this being same as the field produced by/at that point is not very clear to me.
In addition, if we look at the example of two point charges of equal magnitudes but opposite signs kept some distance apart with an infinite plane in between them. Then we can find the force acting between them by integrating Maxwell's Stress Tensor over that whole plane. This is because either of the charges can be considered bounded by a spherical boundary at infinity and the infinite plane. The fields at the spherical boundary is anyways zero so we just integrate over the plane giving the right answer.
How are we, in this case, getting the right answer using the stress tensor approach considering the field created by the point charge is different from the field acting on it (which is created by the second charge) and where both are point charges and not distributions?
 A: You are correct, the standard EM theory based on total field acting on total charge and total current density works when charge and current have finite densities (and even in some cases with singular densities, such as charged surface). This is because contribution of the infinitesimal charged body to the total field is negligible (3D distribution) or finite (2D distribution).
It stops working so well when line or point charges are present, because both total electric field and stress tensor are singular at such charges and the contribution of the infinitesimal charged body is singular. Total force is undefined and total EM energy becomes infinite.
One way to formulate consistent EM theory for such bodies is to assume that there is no self-interaction -- charged bodies, however singular they are, only experience field of other charged bodies. This works even for point charges, as Frenkel has shown, and one can derive appropriate stress tensor for such systems, free of singularities in force or energy [1].
To your example with two point charges, in special cases such as two static charges not on integration surface, the standard "total field stress tensor" approach still works and gives the same result as the more correct Frenkel stress tensor. This is because the two stress tensors differ by a quantity that has zero integral over a closed surface (it is the standard stress tensor of individual electrostatic field of a particle). In general however the two approaches give different results. For example, there is self-interaction of accelerated particles in the standard "total field" tensor theory (the Lorentz-Abraham force that Dirac later derived from standard Poynting-Maxwell stress tensor), but there is no self- force of point particles in Frenkel's theory (there is still a self-force acting on composite bodies, but it is due to interactions between different point particles that compose that body).
[1] Frenkel J., Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. http://dx.doi.org/10.1007/BF01331692
A: The equation $\vec \nabla \cdot \vec E=\rho/\epsilon$ tells you that the divergence of net electric field at a point is equal to the charge density in its neighborhood.
$\vec E\ $  is not the field due to the charge $\rho\ d\tau$ in the infinitesimal volume $d\tau$. It is simply the field at that point due to all charge distributions around.
The charge element $\rho\ d\tau$ inside the volume $\mathcal V$ experiences force due to the electric field $\vec E$ at its location, and that $\vec E$ is in part due to the charge $\rho\ d\tau$ as well, related by the relation $\vec \nabla \cdot \vec E=\rho/\epsilon$.
Note that this can be true only if you assume that the contribution of $\rho\ d\tau$ to the field $\vec E$ is negligible in comparison to the rest of the configuration, so that the field it feels due to others is the same as the total field at its location.
