electrons - Confusion about calculating electrostatic energy using the electric field

Recently I learned that electrostatic potential energy of a system of charges can be calculated like so:

$$E = \int \frac{1}{2} \epsilon_0 \mathbf E^2 dV.$$

However, for a system of two point charges Q1 and Q2, we have

$$E = \int \frac{1}{2} \epsilon_0 (\mathbf E_1 + \mathbf E_2)^2 dV = \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_1^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_2^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 * 2 \mathbf E_1 \cdot \mathbf E_2 dV$$

from which we subtract the first two terms because they are already properties of the charges themselves.

Question:

Why do we not need to do something like this for charge distributions? Aren't charge distributions made up of many many little charges of magnitude $e$? Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?

Answer

Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?

It is instructive to apply a few vector calculus identities to put the electrostatic energy in another form:

$$U = \frac{\epsilon_0}{2} \int E^2 \, d\tau = \frac{\epsilon_0}{2} \int (-\vec{\nabla}V) \cdot \vec{E} \, d \tau = -\frac{\epsilon_0}{2} \oint V \vec{E} \cdot d\vec{a} + \frac{1}{2} \int V (\epsilon_0 \vec{\nabla} \cdot \vec{E}) \, d\tau \\= -\frac{\epsilon_0}{2} \oint V \vec{E} \cdot d\vec{a} + \frac{1}{2} \int V \rho \, d\tau$$ If $\rho$ is non-zero only over a bounded region of space (or in technical terms, $\rho$ has compact support), then it can be shown that the boundary integral vanishes in the limit of an integral over all of space. The second integral, meanwhile, will be finite so long as $\rho$ and $V$ are finite everywhere (since it's effectively an integral of a finite function over a finite region of space.) It's also possible to show that if $\rho$ is finite and of compact support, and if we require $V \to 0$ at infinity, then $V$ will be finite everywhere in space. So if you write this in terms of continuous charge distributions, the electrostatic energy will never diverge.

So that answers your question about whether the integral will diverge. But your question bespeaks a mindset that I would encourage you to move away from as you learn more about E&M: It is better to think of point charges as limits of smooth charge distributions, not the other way around.

It is instructive, when students are first learning E&M, to talk about point charges and infinitesimally thin wires; it's a lot easier to calculate fields, forces, energies, etc. when you only have to do a sum rather than an integral. However, Maxwell's equations are what really govern the behavior of electric and magnetic fields, and these are written in terms of charge and current densities $\rho$ and $\vec{J}$. If you try to model a "point charge" or a "line current" in terms of these quantities, they'll have to be infinite; and whenever you plug in an infinite quantity into an equation, you can't expect it to behave nicely.

Some of the calculations that students do in intro E&M can be rigorously re-derived for point charges by introducing the language of distributions, and in particular the Dirac delta function. You can go a long way by allowing your charge densities, fields, and potentials to be distributions rather than honest-to-god functions. But distributions can't do everything that functions can; and in particular, you're courting danger every time you try to multiply two distributions together, as you have to do to calculate electrostatic energy. Sometimes you can fix this problem with an ad-hoc subtraction (as you found above); sometimes you can't (the Abraham-Lorentz force.)

All of this means that it's better to think of Maxwell's equations as being well-defined only for smooth charge and current densities. So instead of thinking of a smooth charge density as "really" being a bunch of point charges, one should think of a point charge as being a limiting case of a series of smooth charge densities with a fixed total amount of charge as the occupied volume goes to zero. If you have this mentality, then the pathologies that sometimes when solving problems involving infinite charge or current densities will be a lot less confusing.