Dielectrics

Step 1: Potential of a polarized dielectric

A polarized volume produces a potential equivalent to a collection of dipoles. The potential at $\mathbf{r}$ is:

$$\Phi(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int_\mathcal{V}\frac{\mathbf{P}(\mathbf{r}')\cdot(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^3}\,d^3r'$$

Step 2: Use the identity $\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} = \nabla'\frac{1}{|\mathbf{r}-\mathbf{r}'|}$

$$\Phi = \frac{1}{4\pi\epsilon_0}\int\mathbf{P}\cdot\nabla'\frac{1}{|\mathbf{r}-\mathbf{r}'|}\,d^3r'$$

Step 3: Integrate by parts

Using $\nabla'\cdot(\mathbf{P}f) = f\nabla'\cdot\mathbf{P} + \mathbf{P}\cdot\nabla'f$ and the divergence theorem:

$$\Phi = \frac{1}{4\pi\epsilon_0}\left[\oint_\mathcal{S}\frac{\mathbf{P}\cdot\hat{n}'}{|\mathbf{r}-\mathbf{r}'|}\,da' - \int_\mathcal{V}\frac{\nabla'\cdot\mathbf{P}}{|\mathbf{r}-\mathbf{r}'|}\,d^3r'\right]$$

Step 4: Identify bound charges

Comparing with $\Phi = \frac{1}{4\pi\epsilon_0}\int\frac{\rho_{\text{eff}}}{|\mathbf{r}-\mathbf{r}'|}\,d^3r'$, we identify:

$$\sigma_b = \mathbf{P}\cdot\hat{n}, \qquad \rho_b = -\nabla\cdot\mathbf{P}$$

Step 5: Total bound charge is zero

$$Q_b = \oint\sigma_b\,da + \int\rho_b\,d^3r = \oint\mathbf{P}\cdot d\mathbf{a} - \int\nabla\cdot\mathbf{P}\,d^3r = 0$$

by the divergence theorem. Polarization merely redistributes charge; it cannot create net charge.

Step 1: Define atomic polarizability

Each atom develops a dipole moment $\mathbf{p} = \alpha\mathbf{E}_{\text{local}}$ where $\mathbf{E}_{\text{local}}$ is the field actually experienced by the atom (not the macroscopic average field).

Step 2: Compute the local field (Lorentz cavity)

Carve a small spherical cavity around the atom. The local field has three contributions: the external field, the field from bound charges on the cavity surface, and the field from nearby atoms (zero for cubic symmetry):

$$\mathbf{E}_{\text{local}} = \mathbf{E} + \frac{\mathbf{P}}{3\epsilon_0}$$

Step 3: Relate $\mathbf{P}$ to $\mathbf{E}_{\text{local}}$

$$\mathbf{P} = N\alpha\mathbf{E}_{\text{local}} = N\alpha\left(\mathbf{E} + \frac{\mathbf{P}}{3\epsilon_0}\right)$$

Step 4: Solve for $\mathbf{P}$ in terms of $\mathbf{E}$

$$\mathbf{P} = \frac{N\alpha}{1 - N\alpha/(3\epsilon_0)}\mathbf{E} = \epsilon_0\chi_e\mathbf{E}$$

Step 5: Express in terms of $\epsilon_r$

Since $\epsilon_r = 1 + \chi_e$, rearranging gives the Clausius-Mossotti relation:

$$\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{N\alpha}{3\epsilon_0}$$

Note the catastrophe: as $N\alpha/(3\epsilon_0) \to 1$, $\epsilon_r \to \infty$. This signals a ferroelectric transition where spontaneous polarization develops.

Step 1: Energy to polarize a dielectric element

The work to increase the polarization by $d\mathbf{P}$ in a volume element is $dW = -\mathbf{E}\cdot d\mathbf{P}\,d^3r$.

Step 2: Integrate for a linear dielectric

For $\mathbf{P} = \epsilon_0\chi_e\mathbf{E}$, integrating from 0 to the final field:

$$W = \frac{1}{2}\int\mathbf{D}\cdot\mathbf{E}\,d^3r = \frac{1}{2}\int\epsilon|\mathbf{E}|^2\,d^3r$$

Step 3: Force on a dielectric

The force on a dielectric body in a non-uniform field can be found from the energy: $\mathbf{F} = -\nabla W$ at constant charge, or $\mathbf{F} = +\nabla W$ at constant potential.

Step 4: Dielectrophoresis

A dielectric sphere in a non-uniform field experiences a force:

$$\mathbf{F} = 2\pi\epsilon_0 R^3\frac{\epsilon_r - 1}{\epsilon_r + 2}\nabla|\mathbf{E}_0|^2$$

Step 5: Capacitance with dielectrics

For a parallel plate capacitor with dielectric: $C = \epsilon_r\epsilon_0 A/d$. The energy stored is $W = Q^2/(2C)$. Inserting a dielectric at constant charge decreases the energy (the dielectric is pulled in); at constant voltage it increases (external work is done).