The energy stored in a static electric field is calculated as the integral over all space
UE ≡ ε0
2
∫
|E|2 d3x.
For an electric field created by a static charge density ρ(x) with finite extent show that
UE = 1
2
∫
ρ(x)φ(x) d3x
where φ(x) is the electric potential. [Hint: This is an example of using integration by parts in
three dimensions. The finite extent of the charge density means that the potential goes to zero
at infinity at least as fast as r−1. To proceed we recall a couple of facts from electrostatics:
E = −∇φ and ∇ · E = ρ/ε0. Note that |E|2 = E · E, consider expanding ∇ · (ρE), and use
Gauss’ theorem.]

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

The energy stored in a static electric field is calculated as the integral over all space UE ≡ ε0 / 2 ∫ |E|^2 d^3x. For an electric field created by a static charge density ρ(x) with finite extent, show that UE = 1 / 2 ∫ ρ(x)φ(x) d^3x, where φ(x) is the electric potential.

Explore how to express the energy stored in a static electric field using charge density ρ(x) and electric potential φ(x). This problem utilizes key concepts from electrostatics, vector calculus, and Gauss' theorem. Suitable for undergraduate physics or mathematics students.

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