Let us consider the electrostatic potential energy $U_1$ between $n$ point charges $q_1, q_2, q_3$, and so on. Depending upon signs of $q_1, q_2, q_3\dots$, $U_1$ can be both positive or negative. Suppose we have a continuous charge distribution then its Electrostatic Potential Energy $U_2$ is given by half epsilon E (magnitude of Electric Field) squared integrated over whole space; so looking into this expression $U_2$ must always be positive.

$$ U_1>0\text{ or }U_1<0\text{ or }U_1=0\quad\text{but } U_2>0 $$

This means that if the charge distribution is discrete then Electrostatic Potential Energy $U$ can be both positive or negative but for continuous charge distribution Electrostatic Potential Energy $U$ always positive?

Or given any charge distribution what can we say about the general sign of $U$ whether it should be positive or negative as one can easily derive an expression for $U$ for continuous charge distribution from the discrete case expression so what should be the actual sign of $U$?


2 Answers 2


This is a very interesting story which you usually don't get in an undergraduate course on electromagnetism.

For simplicity, consider just two charges $q_1$ at $\mathbf{r}_1$ and $q_2$ at $\mathbf{r}_2$. Their fields are




where I'll use Gaussian units instead of SI units since factors of $\frac{1}{4\pi\epsilon_0}$ are just noise.

The integral for the electrostatic potential energy stored in the combined field $\mathbf{E}_1+\mathbf{E}_2$ is


Expanding the square of the field, there are three terms:





The first two are divergent "self-energy" integrals representing the infinite electrostatic energy of each point charge by itself. They are infinite positive constants that don't depend on the separation of the two charges. They can be ignored; think of them as being a contribution to the mass-energy of the charges, infinitely "renormalizing" their masses, in a classical version of the renormalization you encounter in quantum field theory.

The third, the "interaction-of-$\mathbf{E}_1$-with-$\mathbf{E}_2$" integral, is a finite integral. With some effort it can be performed analytically. (I will not give the details; it's a nontrivial calculation and this answer is already long.) The result turns out to be


which is simply the usual formula for the electrostatic potential energy of two point charges. This integral can be positive or negative depending on the signs of $q_1$ and $q_2$. It isn’t positive-definite because the integrand isn't the square of anything; it’s $\mathbf{E}_1\cdot\mathbf{E}_2$ and dot products can be positive or negative. In fact, in some regions the integrand is positive and in others it’s negative, as you can see by drawing a diagram with some field arrows for the two fields.

So now you know that the usual formula for the electrostatic potential energy of two point charges is the interaction energy of their two individual fields, not the total field energy (which is infinite).

  • $\begingroup$ So why does the continuous charge distribution end up with a positive-definite energy? $\endgroup$
    – Mark H
    Commented Apr 21, 2021 at 7:43

The usual formula for electrostatic potential energy in the discrete case is $$U=\frac{1}{2}\sum_{i}q_iV(\vec{r_i})\tag{1.1},$$ where $V(\vec{r_i})$ is the potential at the position of the $i^{th}$ charge, due to all charges except $q_i$. If $V(\vec{r_i})$ was, instead, the total potential, it would have an infinite value. The analog in the continuous case is $$U = \frac{1}{2}\int{\rho V\,d\tau},\tag{1.2}$$ with the integral carried out over the whole charge distribution (I'm using $\,d\tau$ as the volume element to avoid confusion with the potential, $V$). Now, in keeping with $(1.1)$, $V$ is the potential due to all charges except those within $\,d\tau$. But since the charge inside that volume element is arbitrarily small, its contribution to the potential at its own position is negligible, so we may as well call $V$ the total potential, due to all charges. Some vector calculus quickly yields the formula you referenced: $$U=\frac{1}{2}\epsilon_0\int{E^2\,d\tau}\tag{1.3}.$$ The crucial point is that, to get to this formula, we had to change $V(\vec{r_i})$ (the potential due to all charges except $q_i$) to $V$ (the total potential due to all charges). Conceptually, this means that $(1.1)$ accounts for only the mutual energy of the charges, while $(1.3)$ expresses both the mutual and self-energy of the charge distribution. $(1.1)$ gives the work necessary to assemble a distribution of point charges, ignoring the work to make those charges in the first place. $(1.3)$, on the other hand, is the total work necessary to both make the charges and arrange them. That's why $(1.3)$ always yields a positive value for the energy, while $(1.1)$ can be either positive or negative.


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