While studying about the Maxwell Stress Tensor from Griffiths Chapter 8 Conservation laws, I faced this problem.
First of all, he calculated the force on a volume $\mathcal V$ containing a charge distribution $\rho$ and current density $\vec J$(which were assumed to be finite, since that is the only way IMO you can neglect the fields of infinitesimal volumes when calculating forces on them).
So far so good,the answer came out to be-
$$\vec F= \int_{\mathcal V} \nabla \cdot \overleftrightarrow{T}\ d\tau - \dfrac{d}{dt} \epsilon_o\mu_o\ \int_\mathcal V \ \vec S\ d\tau = \oint_{\mathcal S} \overleftrightarrow{T} \cdot \vec{da} - \dfrac{d}{dt} \epsilon_o\mu_o\ \int_\mathcal V \ \vec S\ d\tau $$
After that, Griffiths makes the claim that integrating over the volume $\mathcal V$ is not necessary, you can take any volume that encloses the whole charge. This claim is obvious, since no charges or currents means no force at those points.
Now, the problem: Let us take two point charges and calculate the Coulombic force between them, using this approach. Consider one charge enclosed within a hemisphere of infinite radius, and you can integrate the stress tensor along the infinite plane between them, the answer comes out to be $q^2/4\pi \epsilon_o r^2$, as expected.
But if the derivation is for finite charge densities, why does an infinite charge density at the location of a charge not ruin the results? My guess is that it is the artistry of converting the volumetric integral into a surface one, which as Griffiths says- sniffs out what is going on inside.
I think that it gets clear precisely because $\int d^3r\delta^3(\vec r) =1$, and this finite integral is subtly involved somewhere which hides the presence of singular charges inside the volume.
Any help is appreciated.
