Electrostatics Review

Step 1: Solid angle subtended by a surface element

For a point charge $q$ at the origin, consider a surface element $d\mathbf{a}$ at position $\mathbf{r}$. The flux through $d\mathbf{a}$ is:

$$d\Phi_E = \mathbf{E}\cdot d\mathbf{a} = \frac{q}{4\pi\epsilon_0}\frac{\hat{r}\cdot d\mathbf{a}}{r^2} = \frac{q}{4\pi\epsilon_0}\,d\Omega$$

where $d\Omega = \hat{r}\cdot d\mathbf{a}/r^2$ is the solid angle subtended by $d\mathbf{a}$ at the charge location.

Step 2: Total flux through a closed surface

Integrating over a closed surface $\mathcal{S}$ enclosing the charge, the total solid angle is $4\pi$:

$$\oint_\mathcal{S} \mathbf{E}\cdot d\mathbf{a} = \frac{q}{4\pi\epsilon_0}\oint d\Omega = \frac{q}{\epsilon_0}$$

If the charge is outside the surface, the net solid angle vanishes, giving zero flux.

Step 3: Generalization by superposition

By superposition, the total flux through any closed surface equals the total enclosed charge divided by $\epsilon_0$:

$$\boxed{\oint_\mathcal{S} \mathbf{E}\cdot d\mathbf{a} = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{1}{\epsilon_0}\int_\mathcal{V} \rho\,d^3r'}$$

Step 4: Differential form via the divergence theorem

Apply the divergence theorem to convert the surface integral to a volume integral:

$$\int_\mathcal{V} \nabla\cdot\mathbf{E}\,d^3r = \frac{1}{\epsilon_0}\int_\mathcal{V} \rho\,d^3r$$

Since this holds for any volume $\mathcal{V}$, the integrands must be equal:

$$\boxed{\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}}$$

Step 5: Connection to the Dirac delta function

Formally, the divergence of the Coulomb field of a point charge is:

$$\nabla\cdot\left(\frac{\hat{r}}{r^2}\right) = 4\pi\delta^3(\mathbf{r})$$

This distributional identity is the mathematical foundation of Gauss's law. It converts the integral statement into a local differential equation.

Step 1: Assume two solutions exist

Suppose $\Phi_1$ and $\Phi_2$ both satisfy $\nabla^2\Phi = -\rho/\epsilon_0$ in $\mathcal{V}$ with $\Phi_1 = \Phi_2$ on $\mathcal{S}$. Define $U = \Phi_1 - \Phi_2$.

Step 2: U satisfies Laplace's equation with zero boundary data

$\nabla^2 U = 0$ in $\mathcal{V}$ and $U = 0$ on $\mathcal{S}$.

Step 3: Apply the identity $\nabla\cdot(U\nabla U) = U\nabla^2 U + |\nabla U|^2$

Integrate over $\mathcal{V}$ and use the divergence theorem:

$$\oint_\mathcal{S} U\frac{\partial U}{\partial n}\,da = \int_\mathcal{V}(U\nabla^2 U + |\nabla U|^2)\,d^3r$$

Step 4: Both terms on left vanish or simplify

Since $U = 0$ on $\mathcal{S}$, the left side vanishes. Since $\nabla^2 U = 0$, we get:

$$\int_\mathcal{V}|\nabla U|^2\,d^3r = 0$$

Step 5: Conclude uniqueness

Since $|\nabla U|^2 \geq 0$ everywhere and its integral vanishes, we must have $\nabla U = 0$ everywhere. Combined with $U = 0$ on the boundary, this gives $U = 0$ throughout $\mathcal{V}$. Therefore $\Phi_1 = \Phi_2$ everywhere. QED.

Step 1: Start with the charge-potential form

$$W = \frac{1}{2}\int\rho\Phi\,d^3r$$

The factor of $1/2$ avoids double-counting pairs of charges.

Step 2: Replace $\rho$ using Gauss's law

Substitute $\rho = \epsilon_0\nabla\cdot\mathbf{E}$:

$$W = \frac{\epsilon_0}{2}\int(\nabla\cdot\mathbf{E})\Phi\,d^3r$$

Step 3: Integration by parts

Use the product rule $\nabla\cdot(\Phi\mathbf{E}) = (\nabla\Phi)\cdot\mathbf{E} + \Phi(\nabla\cdot\mathbf{E})$:

$$W = \frac{\epsilon_0}{2}\left[\oint\Phi\mathbf{E}\cdot d\mathbf{a} - \int(\nabla\Phi)\cdot\mathbf{E}\,d^3r\right]$$

Step 4: Surface term vanishes; use $\mathbf{E} = -\nabla\Phi$

As we integrate over all space, $\Phi$ and $\mathbf{E}$ fall off sufficiently fast that the surface term at infinity vanishes. Since $\nabla\Phi = -\mathbf{E}$:

$$W = \frac{\epsilon_0}{2}\int|\mathbf{E}|^2\,d^3r$$

Step 5: Interpretation

The energy density $u = \frac{\epsilon_0}{2}|\mathbf{E}|^2$ is positive definite, confirming that the electrostatic energy is stored in the field itself. For a point charge, this integral diverges (the self-energy problem), which is an inherent difficulty of classical electrodynamics addressed by renormalization in QED.