electric fields - Electrostatic energy integral for point charges

The electric energy stored in a system of two point charges $Q_1$ and $Q_2$ is simply

$$W = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{a}$$ where $a$ is the distance between them.

However, the total energy can also be calculated through the volume integral of magnitude squared of the electric over all space:

$$W = \frac{\epsilon_0}{2}\int_{\mathbb{R}^3} E^2\,dV$$

Suppose that $Q_1$ sits on the origin and $Q_2$ a distance $a$ away on the $z$-axis. Then the electric field is

$$\vec{E}(x,y,z) = \frac{1}{4\pi\epsilon_0}\left[\left(\frac{Q_1}{r^3} + \frac{Q_2}{d^3}\right)x\hat{x} +\left(\frac{Q_1}{r^3} + \frac{Q_2}{d^3}\right)y\hat{y} + \left(\frac{Q_1z}{r^3} + \frac{Q_2\left(z-a\right)}{d^3}\right)\hat{z}\right]$$ where $$\begin{align} r &= \sqrt{x^2+y^2+z^2} \\ d&= \sqrt{x^2+y^2+\left(z-a\right)^2} \end{align}$$

Calculating $W$ through $\int_{\mathbb{R}^3} E^2\,dV$ seems extremely difficult; Mathematica, for example, appears stumped. Yet its result should simply be $\frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{a}$, correct? The integral formula still applies to point charges, correct?

Answer

Let us call the two electric charges $Q$ and $q$ with electric fields $|{\bf E}_Q|=\frac{k_e |Q|}{r^2_Q}$ and $|{\bf E}_q|=\frac{k_e |q|}{r^2_q}$, respectively. Here $k_e=\frac{1}{4\pi\varepsilon_0}$. The energy

$$\tag{0} U~=~\frac{\varepsilon_0}{2}\iiint_{\mathbb{R}^3}\! d^3r ~|{\bf E}_Q+{\bf E}_q|^2~=~U_Q+U_q+U_{Qq}$$

of the total electric field contains three contributions:

1. 
   

   The energy $U_Q$ of the electric field of a charge $Q$

   $$\tag{1} U_Q~=~\frac{\varepsilon_0}{2}\iiint_{r\geq\delta}\! d^3r ~|{\bf E}_Q|^2~=~ \frac{k_eQ^2}{2} \int_{\delta}^{\infty}\frac{dr}{r^2}~=~\frac{k_eQ^2}{2\delta}.$$
   
   


2. 
   

   The energy $U_q$ of the electric field of a charge $q$

   $$\tag{2} U_q~=~\frac{\varepsilon_0}{2}\iiint_{r\geq\delta}\! d^3r ~|{\bf E}_q|^2~=~ \frac{k_eq^2}{2} \int_{\delta}^{\infty}\frac{dr}{r^2}~=~\frac{k_eq^2}{2\delta}.$$ In equation (1) and (2), we have inserted a regulator $\delta$. If the regulator $\delta\to 0$ is removed the energy becomes infinite.
   
   



3. 
   

   The energy from the exchange term

   $$\tag{3} U_{Qq}~=~\varepsilon_0\iiint_{\mathbb{R}^3}\! d^3r ~{\bf E}_Q\cdot {\bf E}_q~=~\frac{k_eQq}{R},$$ where $R$ is the distance between the two charges. The triple integral (3) can be analytically calculated by (among other things) using the azimuthal symmetry of the integrand.
   
   

If we slowly vary the separation $R$, we will only detect the variation of the exchange term (3) (the Coulomb energy) as the two other contributions (1) and (2) remain fixed.