We are tasked with showing that the energy stored in a static electric field, given by

$U_E = \frac{\varepsilon_0}{2} \int |\mathbf{E}|^2 \, d^3x,$

can be expressed as

$U_E = \frac{1}{2} \int \rho(x) \phi(x) \, d^3x,$

where $\mathbf{E}$ is the electric field, $\phi(x)$ is the electric potential, and $\rho(x)$ is the charge density. We are guided by the relationships $\mathbf{E} = -\nabla \phi$ and $\nabla \cdot \mathbf{E} = \frac{\rho(x)}{\varepsilon_0}$. We will use integration by parts, along with Gauss' theorem, to derive this expression.

Step 1: Express $|\mathbf{E}|^2$ in terms of the potential $\phi$

Recall that $\mathbf{E} = -\nabla \phi$, so the magnitude squared of the electric field is

$|\mathbf{E}|^2 = \mathbf{E} \cdot \mathbf{E} = (-\nabla \phi) \cdot (-\nabla \phi) = (\nabla \phi) \cdot (\nabla \phi).$

Thus, the energy expression becomes

$U_E = \frac{\varepsilon_0}{2} \int (\nabla \phi) \cdot (\nabla \phi) \, d^3x.$

Step 2: Use integration by parts in 3 dimensions

We aim to manipulate this expression using integration by parts. Specifically, we consider the vector identity

$\nabla \cdot (\phi \mathbf{E}) = (\nabla \phi) \cdot \mathbf{E} + \phi (\nabla \cdot \mathbf{E}).$

Substituting $\mathbf{E} = -\nabla \phi$, this becomes

$\nabla \cdot (-\phi \nabla \phi) = -(\nabla \phi) \cdot (\nabla \phi) - \phi \nabla^2 \phi.$

Thus,

$(\nabla \phi) \cdot (\nabla \phi) = -\nabla \cdot (\phi \nabla \phi) - \phi \nabla^2 \phi.$

Substituting this back into the energy expression, we get

$U_E = \frac{\varepsilon_0}{2} \int -\nabla \cdot (\phi \nabla \phi) - \phi \nabla^2 \phi \, d^3x.$

Step 3: Use Gauss' theorem

The first term, $\nabla \cdot (\phi \nabla \phi)$, can be converted into a surface integral using Gauss' theorem:

$\int \nabla \cdot (\phi \nabla \phi) \, d^3x = \int_{\partial V} \phi \nabla \phi \cdot d\mathbf{A}.$

However, we are told that the charge distribution has finite extent, meaning the potential $\phi(x)$ vanishes at infinity (at least as fast as $r^{-1}$), so this surface integral at infinity gives zero. Therefore, the term involving $\nabla \cdot (\phi \nabla \phi)$ vanishes, leaving us with

$U_E = \frac{\varepsilon_0}{2} \int - \phi \nabla^2 \phi \, d^3x.$

Step 4: Use Poisson's equation

From electrostatics, we know that the potential $\phi(x)$ satisfies Poisson's equation:

$\nabla^2 \phi = -\frac{\rho(x)}{\varepsilon_0}.$

Substitute this into the expression for $U_E$:

[ U_E = \frac{\varepsilon_0}{2} \int \phi \frac{\rho(x)}{\varepsilon_0} , d^3x = \frac{1}{2} \int \rho(x) \phi(x) , d^3x