Chapter 6 — Dielectric Matter

Quote

“If we develop only a macroscopic description of matter in an electric field, we shall find it hard to answer some rather obvious sounding questions.”

— Edward M. Purcell (1965)

Motivation

Compared to conductors, dielectrics are unable to completely screen external macroscopic static electric fields from their interior. This is because chemical bonds constrain the internal charge density rearrangements — the same constraints make dielectrics poor conductors of electric current. Physically, we can view conductors as a limiting case of dielectrics in which the non-electrostatic potential energy landscape is perfectly flat.

Polarization — Term Overload

The word “polarization” is used in two distinct senses: (1) the rearrangement of internal charge that occurs when matter is exposed to an external field; (2) the vector field $\mathbf{P}(\mathbf{r})$ used to characterize the details of such rearrangement.

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Free and Bound Charge

Following both historical convention and computational convenience, we denote the charge ${\rho }_f(\mathbf{r})$ originating externally to the dielectric matter as free charge — it is the charge we control experimentally. Its counterpart, the charge responsible for the dielectric-originating field, is usually called bound or polarization charge ${\rho }_P(\mathbf{r})$.

Physically, bound charges arise upon application of an external field: opposite-sign charges are pushed in opposite directions and then rearrange until equilibrium is established — specifically, until ${\rho }_P(\mathbf{r}){\mathbf{E}}_{\text{tot}}(\mathbf{r})$ is equal and opposite to the force density produced by chemical bonds and other non-electrostatic effects.

The total charge density is:

$$\rho (\mathbf{r})={\rho }_f(\mathbf{r})+{\rho }_P(\mathbf{r})$$

It is convenient to split ${\rho }_P$ into volume and surface parts. For an initially neutral dielectric, the bound charge stems from rearrangements (as in a conductor, albeit more constrained), and must integrate to zero:

$${\int }_V{\mathrm{d}}^{3}r{\rho }_P(\mathbf{r})+{\int }_S\mathrm{d}S{\sigma }_P({\mathbf{r}}_S)=0$$

A neutral conductor achieves this with $\rho =0$ everywhere inside and $\sigma \ne 0$ on the surface. For a dielectric, we identify:

$$\begin{array}{rlrl}& {\rho }_P(\mathbf{r})=-\nabla \cdot \mathbf{P}(\mathbf{r})& \mathbf{r}& \in V\\ & {\sigma }_P({\mathbf{r}}_S)=\mathbf{P}({\mathbf{r}}_S)\cdot \hat{\mathbf{n}}({\mathbf{r}}_S)& {\mathbf{r}}_S& \in S\\ & \mathbf{P}(\mathbf{r})=0& \mathbf{r}& \notin V\end{array}$$

The last follows because $\mathbf{P}$ is associated with matter and thus confined to the sample.

Surface Terms from Theta Functions

It is always possible to write ${\rho }_P$ as a single density using step functions. For example, if $\mathbf{P}(\mathbf{r})={\mathbf{P}}_0(\mathbf{r})\Theta (z)$, then:

$${\rho }_P=-\nabla \cdot \mathbf{P}(\mathbf{r})=-\Theta (z)\nabla \cdot {\mathbf{P}}_0-\delta (z)P_{0z}$$

The last term is the surface contribution.

Non-Uniqueness of $\mathbf{P}$

The three defining equations above do not determine $\mathbf{P}$ uniquely, precisely because $\nabla \times \mathbf{P}(\mathbf{r})$ need not vanish inside $V$. By the Helmholtz decomposition, knowing $\nabla \cdot \mathbf{P}$ only fixes the longitudinal (irrotational) part; the transverse (solenoidal) part remains free.

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The Dipole Moment of Polarized Matter

Using the identity $\nabla \cdot (r_k\mathbf{P})=P_k+r_k\nabla \cdot \mathbf{P}$ and integrating over the sample volume:

$${\int }_V{\mathrm{d}}^{3}r\mathbf{P}(\mathbf{r})=-{\int }_V{\mathrm{d}}^{3}r\mathbf{r}(\nabla \cdot \mathbf{P})+{\int }_S\mathrm{d}S(\hat{\mathbf{n}}\cdot \mathbf{P})\mathbf{r}={\int }_V{\mathrm{d}}^{3}r\mathbf{r}{\rho }_P(\mathbf{r})+{\int }_S\mathrm{d}S{\sigma }_P({\mathbf{r}}_S){\mathbf{r}}_S=\mathbf{p}$$

Alternative Derivation (Without Splitting into Volume and Surface)

Start from the Coulomb integral for the potential produced by polarization charges. Far from the sample, expand $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ in a multipole series. The dipole term is ${\phi }_{\text{dip}}=\mathbf{p}\cdot \hat{\mathbf{r}}/(4\pi {ϵ}_0{r}^2)$. Comparing with the potential produced by a volume distribution of dipoles $\mathrm{d}\mathbf{p}=\mathbf{P}{\mathrm{d}}^{3}r$ (derived below), we read off $\mathbf{p}={\int }_V{\mathrm{d}}^{3}r\mathbf{P}(\mathbf{r})$ directly, without ever separating volume and surface contributions.

With simple Lorentz averaging of atomic/molecular dipole moments, the previous integral tempts us to identify $\mathbf{P}(\mathbf{r})$ as the electric dipole moment per unit volume, and — like with point charges — simply superpose a collection of atomic/molecular dipoles.

Why Lorentz Averaging Fails

Despite the clean result ${\int }_V{\mathrm{d}}^{3}r\mathbf{P}=\mathbf{p}$, the Lorentz average:

$$\mathbf{P}(\mathbf{R})=\frac{1}{\Omega }{\int }_{\Omega }{\mathrm{d}}^{3}s\mathbf{s}{\rho }_{\text{micro}}(\mathbf{s})=\frac{\mathbf{p}(\mathbf{R})}{\Omega }$$

is usually a poor approximation to the true polarization, except for dielectric gases, non-polar liquids, and molecular solids where the constituent atoms and molecules interact very weakly. The failure has two origins:

1. Bond charge: Chemical bonding places significant charge on the boundaries of the averaging cell $\Omega$, which is not captured by treating each cell as an isolated entity.
2. Cell dependence: Different but equally sensible choices of $\Omega$ yield different values of $\mathbf{P}$ — one can even choose $\Omega$ so that $\mathbf{P}=0$.

TODO: Add a figure showing completely opposite $\mathbf{P}$ from different averaging cell choices, and a choice giving $\mathbf{P}=0$.

The Modern Theory of Polarization

The essential insight (Resta and Vanderbilt, 2007) resolves a paradox: ${\rho }_P=-\nabla \cdot \mathbf{P}$ implies that $\mathbf{P}$ contains more information than ${\rho }_P$ (a vector vs. a scalar, and many different $\mathbf{P}$s can produce the same ${\rho }_P$). But if $\mathbf{P}$ is simply the dipole moment per unit volume, it contains less information than $\rho (\mathbf{r})$ (it captures only the smoothly varying dipole orientations, missing high-frequency charge density fluctuations).

The resolution lies in quantum mechanics: $\rho (\mathbf{r})$ is determined by $|\Psi {|}^2$, while $\mathbf{P}(\mathbf{r})$ encodes information about the phase of the system wavefunction.

Recalling that the polarization current density ${\mathbf{j}}_P=\partial \mathbf{P}/\partial t$ is associated with time variations of polarization, we can attempt to upgrade Lorentz averaging:

$$\mathbf{P}(\mathbf{R})=\frac{1}{\Omega }{\int }_{\Omega }{\mathrm{d}}^{3}s{\int }_{-\infty }^{\infty }\mathrm{d}t{\mathbf{j}}_{P,\text{micro}}(\mathbf{s},t)$$

This asks: how much did the charges move in total when we turn on the external field? The integration from $-\infty$ to $\infty$ covers the full history of the polarization process.

NaBr Crystal Example

Quantum-mechanical calculation of ${\mathbf{j}}_{P,\text{micro}}$ for NaBr crystals shows that the usual Lorentz-averaged $\mathbf{P}(\mathbf{r})$ accounts for less than half of the true polarization found from the current-based expression. The majority of $\mathbf{P}(\mathbf{r})$ stems from polarization currents that flow between averaging cells. For covalent dielectrics like silicon, interpreting polarization in terms of cell dipole moments is completely qualitatively incorrect.

The punchline: Lorentz’s model is realistic only when entities like polarized atoms/molecules retain their individual integrity as quantum objects (gases, simple liquids, and van der Waals–bonded solids satisfy this, but the vast majority of dielectrics do not), and we need more sophisticated methods to calculate $\mathbf{P}(\mathbf{r})$. However, once we know $\mathbf{P}(\mathbf{r})$ by whatever means, Lorentz’s notion that a dielectric behaves like a continuous distribution of point electric dipoles with density $\mathbf{P}(\mathbf{r})$ is rigorously correct, as we will now show.

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Fields Produced by Polarized Matter

Suppose $\mathbf{P}(\mathbf{r})$ is known. The potential and field produced by the polarization charge densities are:

$${\phi }_P(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^3{r}^{\prime }\frac{-{\nabla }^{\prime }\cdot \mathbf{P}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}+\frac{1}{4\pi {ϵ}_0}{\int }_S\frac{\mathrm{d}{\mathbf{S}}^{\prime }\cdot \mathbf{P}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$ $${\mathbf{E}}_P(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^3{r}^{\prime }\frac{-{\nabla }^{\prime }\cdot \mathbf{P}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}(\mathbf{r}-{\mathbf{r}}^{\prime })+\frac{1}{4\pi {ϵ}_0}{\int }_S\frac{\mathrm{d}{\mathbf{S}}^{\prime }\cdot \mathbf{P}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}(\mathbf{r}-{\mathbf{r}}^{\prime })$$

These are exact expressions, convergent and valid at all points in space.

The Point Dipole Representation

Rewriting the surface integral using the divergence theorem and integrating by parts to cancel volume terms:

$${\phi }_P(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^3{r}^{\prime }\mathbf{P}({\mathbf{r}}^{\prime })\cdot {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Recalling the potential of a point dipole $\phi (\mathbf{r})=-\frac{1}{4\pi {ϵ}_0}\mathbf{p}\cdot \nabla \frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$ and using ${\nabla }^{\prime }\to -\nabla$, we see that the electrostatic potential is that of a collection of point electric dipoles with moments $\mathrm{d}\mathbf{p}({\mathbf{r}}^{\prime })={\mathrm{d}}^3{r}^{\prime }\mathbf{P}({\mathbf{r}}^{\prime })$.

Physical Significance

Although there are no point dipoles in nature and Lorentz’s averaging is not correct per se, his idea to represent a polarized solid by a volume distribution of point dipoles is valid — it’s just that we need a better method than Lorentz averaging to obtain $\mathbf{P}(\mathbf{r})$.

This view also gives intuition for the physical origin of bound charges: a row of aligned dipoles ($+-+-+-$) cancels in the bulk (unless $\mathbf{P}$ varies spatially, as measured by $\nabla \cdot \mathbf{P}$), while uncompensated charges are left sticking out at the surface.

The Naïve Dipole Formula Diverges Inside $V$

The expression:

$${\mathbf{E}}_P^{†}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_V{\mathrm{d}}^3{r}^{\prime }\left[\frac{3(\mathbf{r}-{\mathbf{r}}^{\prime })[\mathbf{P}({\mathbf{r}}^{\prime })\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })]}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^5}-\frac{\mathbf{P}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}\right]$$

agrees with ${\mathbf{E}}_P$ for $\mathbf{r}\notin V$ but diverges for $\mathbf{r}\in V$. The divergence can be handled by scooping out an infinitesimal vacuum cavity around $\mathbf{r}$ and studying the vanishing-size limit, but the result depends on the shape of the cavity. We dispense with this approach entirely.

Poisson’s Formula for Uniform Polarization

Poisson's Formula

When $\mathbf{P}$ is uniform, $\nabla \cdot \mathbf{P}=0$ inside $V$ and only the surface term survives. Manipulating ${\nabla }_k^{\prime }(P_k(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3)$, pulling the constant $P_k$ outside the integral, and using ${\nabla }_k^{\prime }\to -{\nabla }_k$:

$${\mathbf{E}}_P(\mathbf{r})=-(\mathbf{P}\cdot \nabla )\mathcal{E}(\mathbf{r})$$

where $\mathcal{E}(\mathbf{r})$ is the electric field produced by the same shaped object filled with a uniform charge density of unit magnitude. This reduces the problem of a uniformly polarized body to a standard electrostatics problem.

Application — A Uniformly Polarized Sphere

The electric field of a sphere of radius $R$ with uniform polarization $\mathbf{P}$ plays a role in many applications. We derive it by two methods.

Method I (Poisson’s formula): The electric field of a uniformly charged sphere ($\rho =1$) is:

$$\mathcal{E}(\mathbf{r})=\{\begin{array}{ll}\frac{\mathbf{r}}{3{ϵ}_0}& r<R\\ \frac{V\hat{\mathbf{r}}}{4\pi {ϵ}_0{r}^2}& r>R\end{array}$$

Using ${\nabla }_i{r}_j={\delta }_{ij}$ and ${\nabla }_i(r_j/r^3)={\nabla }_j(r_i/r^3)$, Poisson’s formula gives:

$${\mathbf{E}}_P(\mathbf{r})=\{\begin{array}{ll}-\frac{\mathbf{P}}{3{ϵ}_0}& r<R\\ \frac{V}{4\pi {ϵ}_0}\frac{3(\hat{\mathbf{r}}\cdot \mathbf{P})\hat{\mathbf{r}}-\mathbf{P}}{r^3}& r>R\end{array}$$

Inside the sphere, ${\mathbf{E}}_P$ is constant and anti-parallel to $\mathbf{P}$. Outside, it is exactly the field of a point dipole $\mathbf{p}=V\mathbf{P}$. Comparing the interior field with the delta-function part of the point dipole field (Chapter 4), we conclude that a point electric dipole may be regarded as the $R\to 0$ limit of a uniformly polarized sphere.

Method II (Surface charge): For $\mathbf{P}=P\hat{\mathbf{z}}$, the surface polarization charge density is ${\sigma }_P=\mathbf{P}\cdot \hat{\mathbf{r}}=P\cos \theta$. In Application 4.3, a charge density $\sigma \propto \cos \theta$ on a sphere produces potentials varying as $r\cos \theta$ inside and $\cos \theta /r^2$ outside:

$$\phi (r,\theta )=\frac{P}{3{ϵ}_0}\{\begin{array}{ll}z& r<R\\ \frac{R^3}{r^2}\cos \theta & r>R\end{array}$$

Computing $\mathbf{E}=-\nabla \phi$ reproduces the result above.

TODO: Add a sketch of the field lines (program this).

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The Total Electric Field and the Displacement Field

Using the definition of bound charge and Gauss’s law:

$${ϵ}_0\nabla \cdot \mathbf{E}(\mathbf{r})=\rho (\mathbf{r})={\rho }_f(\mathbf{r})+{\rho }_P(\mathbf{r})={\rho }_f(\mathbf{r})-\nabla \cdot \mathbf{P}(\mathbf{r})$$

Both for historical reasons and computational convenience, we define the auxiliary field (or electric displacement):

$$\mathbf{D}(\mathbf{r}):={ϵ}_0\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})$$

This satisfies:

$$\nabla \cdot \mathbf{D}={\rho }_f,\nabla \times \mathbf{D}=\nabla \times \mathbf{P}$$

The displacement field allows us to apply the usual electrostatic machinery to free charges alone — convenient since ${\rho }_f$ is what we control experimentally, while ${\rho }_P$ is usually a response to ${\rho }_f$.

Gauss’s law for free charge:

$${\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{D}=Q_{f,\text{enc}}$$

Matching conditions at an interface with outward normal ${\hat{\mathbf{n}}}_2$ pointing from medium 1 into medium 2:

$${\hat{\mathbf{n}}}_2\cdot ({\mathbf{D}}_1-{\mathbf{D}}_2)={\sigma }_f,{\hat{\mathbf{n}}}_2\times ({\mathbf{D}}_1-{\mathbf{D}}_2)={\hat{\mathbf{n}}}_2\times ({\mathbf{P}}_1-{\mathbf{P}}_2)$$

$\mathbf{D}$ Is Not Determined by Free Charge Alone

Since $\nabla \times \mathbf{D}\ne 0$ in general, by the Helmholtz decomposition free charge is not the sole source of $\mathbf{D}$. Spatial variations of $\mathbf{P}$ also contribute:

$$\mathbf{D}(\mathbf{r})=-\nabla \int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_f({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}+\nabla \times \int {\mathrm{d}}^3{r}^{\prime }\frac{{\nabla }^{\prime }\times \mathbf{P}({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

The Need for a Constitutive Relation

We have three fields $\mathbf{D}$, $\mathbf{E}$, $\mathbf{P}$ related by $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$, and two curl/divergence equations: $\nabla \times \mathbf{E}=0$ and $\nabla \cdot \mathbf{D}={\rho }_f$. This system is underdetermined without either specifying $\mathbf{P}(\mathbf{r})$ directly (as in ferroelectrics or piezoelectrics, where $\mathbf{P}$ is fixed, not a response to the field) or providing a constitutive relation connecting $\mathbf{P}$ and $\mathbf{E}$ — an equation that imbues electromagnetism with actual properties of dielectric matter (from quantum mechanics or statistical mechanics).

Experiments show that in general:

$$P_i={ϵ}_0{\chi }_{ij}{E}_j+{ϵ}_0{\chi }_{ijk}^{(2)}{E}_j{E}_k+\cdots$$

The $\chi$ are tensors, so $\mathbf{P}$ need not be parallel to $\mathbf{E}$. All matter exhibits nonlinear response for large enough electric field strengths.

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Simple Dielectric Matter

Restricting to linear and spatially isotropic response defines a simple dielectric:

$$\mathbf{P}={ϵ}_0\chi \mathbf{E}$$

where $\chi$ is the electric susceptibility. Defining the permittivity $ϵ$ and dielectric constant $\kappa$:

$$\mathbf{P}=(ϵ-{ϵ}_0)\mathbf{E}={ϵ}_0(\kappa -1)\mathbf{E},\mathbf{D}=ϵ\mathbf{E}=\kappa {ϵ}_0\mathbf{E}={ϵ}_0(1+\chi )\mathbf{E}$$

so $ϵ={ϵ}_0(1+\chi )$ and $\kappa =1+\chi$.

Locality and Positivity

• Locality: These formulas apply pointwise — $\mathbf{P}(\mathbf{r})={ϵ}_0\chi \mathbf{E}(\mathbf{r})$. In general, $\mathbf{P}$ could depend on $\mathbf{E}$ in a neighborhood: $\mathbf{P}(\mathbf{r})={ϵ}_0\int {\mathrm{d}}^3{r}^{\prime }\chi (\mathbf{r},{\mathbf{r}}^{\prime })\mathbf{E}({\mathbf{r}}^{\prime })$, but for simple media this dependence is local, which makes textbook problems solvable.
• Positivity: $\chi \ge 0$, implying $ϵ\ge {ϵ}_0$ and $\kappa \ge 1$. (See T.M. Sanders, Jr., “On the sign of the static susceptibility”, American Journal of Physics 56, 448 (1988).)

Example — Dipole Screening in a Dielectric Sphere

A point dipole ${\mathbf{p}}_0$ is embedded at the center of a dielectric sphere with volume $V$ and dielectric constant $\kappa$. Find the total dipole moment ${\mathbf{p}}_{\text{tot}}$.

From ${\int }_V{\mathrm{d}}^{3}r\mathbf{P}=\mathbf{p}$ and $\mathbf{P}={ϵ}_0(\kappa -1)\mathbf{E}$:

$$\mathbf{p}={ϵ}_0(\kappa -1){\int }_V{\mathrm{d}}^{3}r\mathbf{E}$$

From Example 4.1, ${\int }_V{\mathrm{d}}^{3}r\mathbf{E}=-{\mathbf{p}}_{\text{tot}}/(3{ϵ}_0)$. Therefore:

$${\mathbf{p}}_{\text{tot}}={\mathbf{p}}_0+\mathbf{p}={\mathbf{p}}_0-\frac{1}{3}(\kappa -1){\mathbf{p}}_{\text{tot}}⟹{\mathbf{p}}_{\text{tot}}=\frac{3}{2+\kappa }{\mathbf{p}}_0$$

Since $\kappa \ge 1$, the dielectric medium generally screens (reduces the magnitude of) the embedded dipole.

Bound Charge in Simple Matter

Analyzing $\nabla \cdot \mathbf{D}={\rho }_f$ with $\mathbf{D}=ϵ\mathbf{E}$:

$$ϵ\nabla \cdot \mathbf{E}+\mathbf{E}\cdot \nabla ϵ={\rho }_f$$

When $ϵ$ varies in space, this is a difficult differential equation. We restrict attention to regions where $ϵ$ is piecewise uniform, so that in each region:

$$ϵ{\nabla }^2\phi =-{\rho }_f$$

(valid since $\nabla \times \mathbf{E}=0$ always in electrostatics). Using $\Theta$-functions, one can write $ϵ$ valid everywhere and integrate over boundaries to recover the matching conditions.

Surface Contributions to $\mathbf{D}$ in Simple Matter

For simple matter, $\mathbf{P}\propto \mathbf{E}$ so $\nabla \times \mathbf{P}=0$ in each uniform region. The curl integral in the Helmholtz decomposition of $\mathbf{D}$ seems to vanish. But $ϵ$ jumps at interfaces: writing $ϵ$ with $\Theta$-functions gives $\nabla \times (\mathbf{P}\Theta )\sim \mathbf{P}\times \hat{\mathbf{n}}\delta (\text{surface})$, collapsing the volume integral to a surface one:

$$\mathbf{D}(\mathbf{r})=-\nabla \int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_f({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}+\sum _k\nabla \times {\int }_{S_k}\frac{\mathbf{P}({\mathbf{r}}^{\prime })\times \hat{\mathbf{n}}({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}\mathrm{d}{S}^{\prime }$$

The volume and surface polarization charges in simple matter take convenient forms:

$${\rho }_P=-\nabla \cdot \mathbf{P}=\left(\frac{1}{\kappa }-1\right){\rho }_f,{\sigma }_P=\mathbf{P}\cdot \hat{\mathbf{n}}{|}_S={ϵ}_0(\kappa -1)\mathbf{E}\cdot \hat{\mathbf{n}}{|}_S$$

The first formula gives the total charge density:

$$\rho ={\rho }_P+{\rho }_f=\frac{{\rho }_f}{\kappa }$$

This is the macroscopic screening effect: the induced polarization charge occupies the same location as the free charge, partially canceling it. For a point charge $q$ embedded in a simple dielectric medium:

$$\mathbf{E}(\mathbf{r})=\frac{{\mathbf{E}}_{\text{vac}}(\mathbf{r})}{\kappa },\phi (\mathbf{r})=\frac{{\phi }_{\text{vac}}(\mathbf{r})}{\kappa }$$

Screening Holds Everywhere

This screening of field and potential holds for observation points both inside and outside the dielectric medium — a consequence of $\rho (\mathbf{r})={\rho }_f(\mathbf{r})/\kappa$ being correct pointwise.

Parallel-Plate Capacitor Examples

Fixed Charge (Isolated Plates)

Start with vacuum capacitance $C_0={ϵ}_{0}A/d$ and surface charge density ${\sigma }_f=Q_f/A$. After inserting a dielectric, the charge is fixed (plates are isolated). Gauss’s law (with no fields outside) gives $D={\sigma }_f\hat{\mathbf{z}}$, so:

$$E=\frac{D}{ϵ}=\frac{E_0}{\kappa }$$

The field is screened, consistent with free charges being partially neutralized by polarization charges: ${\sigma }_P=-\mathbf{P}\cdot \hat{\mathbf{z}}=\frac{1-\kappa }{\kappa }{\sigma }_f$, giving total surface charge $\sigma ={\sigma }_f+{\sigma }_P={\sigma }_f/\kappa$.

The potential difference scales by $1/\kappa$, so:

$$C=\kappa C_0$$

Physical meaning: Inserting a dielectric into a capacitor increases its capacitance — it can store more charge at a given voltage. This is used extensively in electronics to make compact, high-capacitance components.

Fixed Potential (Battery-Connected Plates)

Now the plates are connected to a battery maintaining potential $V$. The field is fixed: $E=E_0=V/d$, but $D={ϵ}_0\kappa E$, so ${\sigma }_f=D\cdot \hat{\mathbf{z}}={ϵ}_0\kappa V/d$. The polarization charge on the lower plate is ${\sigma }_P=-\mathbf{P}\cdot \hat{\mathbf{z}}=-{ϵ}_0(\kappa -1)V/d$.

Again $C=\kappa C_0$.

Physical picture: The battery lets additional free charge flow onto the plates to cancel the polarization charge and retain the original field/potential. The charge densities sum to keep the field unchanged. Faraday used the relation ${\sigma }_f={ϵ}_0\kappa V/d$ to measure $\kappa$ for several materials, observing the change in charge drawn onto the plates of a spherical capacitor upon insertion of a dielectric.

Microscopic vs. Macroscopic Picture

Macroscopically, the free and polarization charge densities appear coincident on the plate surfaces. Microscopically, this is not the case — the field immediately adjacent to both metal plates is larger in magnitude than the vacuum field, while there are regions inside the dielectric where it is smaller (or oppositely directed). This connects to the discussion of the local field in the Clausius–Mossotti theory below.

Charge at a Dielectric Interface

A Point Charge in Nested Dielectrics

A point charge $q$ is embedded in a sphere of radius $R$ with dielectric constant ${\kappa }_1$, itself embedded in an infinite medium with dielectric constant ${\kappa }_2$. By spherical symmetry and Gauss’s law for free charge:

$$\mathbf{D}(\mathbf{r})=\frac{q}{4\pi }\frac{\hat{\mathbf{r}}}{r^2}$$

everywhere. Since $\mathbf{D}=ϵ\mathbf{E}$: ${\mathbf{E}}_1=q\hat{\mathbf{r}}/(4\pi {ϵ}_1{r}^2)$ in medium 1 and ${\mathbf{E}}_2=q\hat{\mathbf{r}}/(4\pi {ϵ}_2{r}^2)$ in medium 2.

Apparent paradox: The screened field ${\mathbf{E}}_q=q\hat{\mathbf{r}}/(4\pi {ϵ}_1{r}^2)$ holds everywhere in space. How does ${\mathbf{E}}_2$ arise in medium 2?

Resolution: ${\mathbf{E}}_2={\mathbf{E}}_q+{\mathbf{E}}_{\sigma }$, where ${\mathbf{E}}_{\sigma }$ is the field from a uniform surface polarization charge ${\sigma }_P$ at the $r=R$ interface. By Gauss’s law:

$${\mathbf{E}}_{\sigma }(\mathbf{r})=\frac{{\sigma }_P{R}^2}{{ϵ}_0}\frac{\hat{\mathbf{r}}}{r^2},r>R$$

Both dielectrics contribute to ${\sigma }_P$: ${\sigma }_P=\hat{\mathbf{r}}\cdot ({\mathbf{P}}_1-{\mathbf{P}}_2){|}_{r=R}$. Since ${\mathbf{D}}_1(R)={\mathbf{D}}_2(R)$ (no free charge at the interface):

$${\sigma }_P=-{ϵ}_0\hat{\mathbf{r}}\cdot \mathbf{D}(R)\left(\frac{1}{{ϵ}_1}-\frac{1}{{ϵ}_2}\right)$$

Substituting back confirms ${\mathbf{E}}_2={\mathbf{E}}_q+{\mathbf{E}}_{\sigma }=q\hat{\mathbf{r}}/(4\pi {ϵ}_2{r}^2)$.

Why $\mathbf{D}$ Is Simple Here In this problem, there is no subtlety with $\nabla \times \mathbf{P}$ or surface contributions to $\mathbf{D}$: the spherical symmetry ensures $\mathbf{P}\times \hat{\mathbf{n}}=0$ at the interface (both $\mathbf{P}$ and $\hat{\mathbf{n}}$ are radial). The displacement field is entirely determined by $q$ via Gauss's law, making $\mathbf{D}$ the natural quantity to work with. The bound charges at the interface alter $\mathbf{E}$ and $\mathbf{P}$, but $\mathbf{D}$ is oblivious to them.

A Point Charge Between Two Semi-Infinite Dielectrics

Charge $q$ at $z=-d$; half-spaces ${\kappa }_L$ ($z<0$) and ${\kappa }_R$ ($z>0$). (For the image-charge solution, see Section 8.3.3.)

Strategy: The normal field at the interface has two sources: the screened point charge $q/{\kappa }_L$ and the surface polarization charge ${\sigma }_P$ itself. The latter contributes $\pm {\sigma }_P/2$ on opposite sides (infinite-sheet result). Imposing ${\kappa }_L{E}_{Lz}={\kappa }_R{E}_{Rz}$ (continuity of $D_z$, no free surface charge) is a single equation for ${\sigma }_P$:

$${\sigma }_P(x,y)=\frac{1}{2\pi {\kappa }_L}\frac{{\kappa }_L-{\kappa }_R}{{\kappa }_L+{\kappa }_R}\frac{qd}{(x^2+y^2+d^2{)}^{3/2}}$$

Key lesson: The general Helmholtz formula for $\mathbf{D}$ requires knowing $\mathbf{P}$ everywhere — when $\mathbf{P}$ is itself the unknown, it leads to an integral equation. Here the self-consistent approach via matching conditions is far simpler.

Refraction at a Dielectric Interface

Law of Dielectric Refraction

At a flat interface between media with permittivities ${ϵ}_1$ and ${ϵ}_2$ (no free surface charge), the continuity of normal $\mathbf{D}$ and tangential $\mathbf{E}$ gives:

$${ϵ}_1{E}_1\cos {\alpha }_1={ϵ}_2{E}_2\cos {\alpha }_2,E_1\sin {\alpha }_1=E_2\sin {\alpha }_2$$

where ${\alpha }_k$ is the angle between ${\mathbf{E}}_k$ and the interface normal. Dividing:

$${ϵ}_1\cot {\alpha }_1={ϵ}_2\cot {\alpha }_2$$

Interesting limits: When ${ϵ}_2\gg {ϵ}_1$, both ${\alpha }_1\approx 0$ and ${\alpha }_2\approx \pi /2$ satisfy the law. The first is consistent with $\mathbf{E}=(\sigma /{ϵ}_0)\hat{\mathbf{n}}$ just outside a perfect conductor. The second says that lines of $\mathbf{D}$ and $\mathbf{E}$ in the high-permittivity material are nearly parallel to the interface with the low-permittivity material. Which solution Nature adopts depends on the detailed geometry and the positions of free charges away from the interface.

Applications: This refraction law governs field distributions in layered dielectric structures (capacitors, insulation, geological strata), and is important in the design of high-voltage equipment where field concentration at interfaces can cause dielectric breakdown.

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Potential Theory for Simple Matter

In each uniform region: $ϵ\nabla \cdot \mathbf{E}={\rho }_f$, or equivalently:

$${\nabla }^2\phi =-\frac{{\rho }_f}{\kappa {ϵ}_0}$$

with matching conditions at interfaces:

$${\phi }_1({\mathbf{r}}_S)={\phi }_2({\mathbf{r}}_S),\left({\kappa }_2\frac{\partial {\phi }_2}{\partial n_2}-{\kappa }_1\frac{\partial {\phi }_1}{\partial n_2}\right){|}_S=\frac{{\sigma }_f}{{ϵ}_0}$$

When there are no free charges anywhere, this reduces to Laplace’s equation in each region with:

$${\kappa }_1\frac{\partial {\phi }_1}{\partial n}{|}_S={\kappa }_2\frac{\partial {\phi }_2}{\partial n}{|}_S$$

General solution techniques are developed in Chapter 7 (separation of variables) and Chapter 8 (Green functions).

Example — Spherical Cavity in a Dielectric

Spherical vacuum cavity (radius $R$) in an infinite medium ($\kappa$) with applied uniform field ${\mathbf{E}}_0$. Surface charge ${\sigma }_P\propto \cos \theta$ dictates the $\ell =1$ Legendre structure. Matching $\phi$ and $\kappa \partial \phi /\partial r$ at $r=R$:

$${\mathbf{E}}_{\text{in}}=\frac{3\kappa }{2\kappa +1}{\mathbf{E}}_0>{\mathbf{E}}_0$$

The cavity field is enhanced — bound charges on the cavity wall reinforce ${\mathbf{E}}_0$ (positive charge on the upstream hemisphere, negative downstream). Contrast with the conducting sphere, where the interior field vanishes.

Self-consistency subtlety: The surface charge used to motivate the $\cos \theta$ ansatz differs from the final self-consistent ${\sigma }_P$, but both share the same angular dependence — the matching conditions fix the magnitudes.

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Physics of the Dielectric Constant

Substance	$\kappa$
He	1.000065
N${}_2$	1.00055
CH${}_4$	1.7
SiO${}_2$	4.5
Si	11.8
H${}_2$O	80

Calculating these values requires understanding the local electric field that exists inside a polarized dielectric.

Polarizability

From the uniformly polarized sphere result, we introduce the polarizability $\alpha$ of a small dielectric body as the proportionality between its induced dipole moment and the uniform external field:

$$\mathbf{p}=\alpha {ϵ}_0{\mathbf{E}}_0$$

For a dielectric sphere: $\mathbf{p}=4\pi {ϵ}_0{R}^3{\mathbf{E}}_0=3V{ϵ}_0{\mathbf{E}}_0$, so $\alpha =3V$. This extends to atomic/molecular scales: $\alpha$ can be estimated from $\alpha \approx 3V_{\text{molecule}}$.

Clausius–Mossotti Relation

The macroscopic field $\mathbf{E}(\mathbf{R})=\frac{1}{\Omega }{\int }_{\Omega }{\mathrm{d}}^{3}r{\mathbf{E}}_{\text{micro}}(\mathbf{r})$ is the average of the field in a given cell from all sources, including the cell’s own (self) field. But polarization measures the response to the external field, so we exclude the self-field:

$$\mathbf{P}(\mathbf{R})=n\alpha {ϵ}_0{\mathbf{E}}_{\text{ext}}(\mathbf{R})=n\alpha {ϵ}_0\left(\mathbf{E}(\mathbf{R})-\frac{1}{\Omega }{\int }_{\Omega }{\mathrm{d}}^{3}s{\mathbf{E}}_{\text{self}}(\mathbf{s})\right)$$

where $n$ is the number density of polarizable entities. From the average-field-inside-a-sphere result (Chapter 4):

$$\frac{1}{\Omega }{\int }_{\Omega }{\mathrm{d}}^{3}s{\mathbf{E}}_{\text{self}}(\mathbf{s})=-\frac{\mathbf{p}}{3{ϵ}_0\Omega }=-\frac{\mathbf{P}(\mathbf{R})}{3{ϵ}_0}$$

(exact for a spherical molecule; a good approximation for non-spherical cells when the cell is large compared to the molecular charge distribution). This defines the local field — the actual field experienced by each molecule:

$${\mathbf{E}}_{\text{local}}=\mathbf{E}(\mathbf{R})+\frac{\mathbf{P}(\mathbf{R})}{3{ϵ}_0}=\frac{\kappa +2}{3}\mathbf{E}(\mathbf{R})$$

The local field exceeds the average macroscopic field — each molecule sits in a cavity where the surrounding polarization enhances the field.

Substituting back:

$$\mathbf{P}(\mathbf{R})=\frac{n\alpha {ϵ}_0}{1-n\alpha /3}\mathbf{E}(\mathbf{R})$$

Using $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ gives $\kappa =1+n\alpha /(1-n\alpha /3)$, usually written as the Clausius–Mossotti formula:

$$\frac{\kappa -1}{\kappa +2}=\frac{n\alpha }{3}$$

Dilute Limit

When $n\alpha \ll 1$: $\mathbf{P}\approx n\alpha {ϵ}_0\mathbf{E}$, so $\kappa \approx 1+n\alpha$. The Clausius–Mossotti correction becomes important when the molecules are dense enough that the local field deviates significantly from the macroscopic field.

Non-polar gases and some simple liquids (He, N${}_2$, C${}_6$H${}_6$) obey Clausius–Mossotti well. CH${}_4$ also obeys it — but H${}_2$O does not. The reason is that methane is a non-polar molecule, while water has a permanent dipole moment ${\mathbf{p}}_0$. Onsager realized that ${\mathbf{p}}_0$ modifies the local field needed to calculate $\kappa$, and derived a generalization of Clausius–Mossotti using a cavity approach that includes ${\mathbf{p}}_0$ at the center (see Chapter 8).

For solids such as SiO${}_2$ and Si, there is no simple formula — quantum-mechanical perturbation theory is required.

The Conductor Limit

As $\kappa \to \infty$: either from Clausius–Mossotti ($n\alpha \to 3$) or from the relation between $\mathbf{E}(\mathbf{R})$ and $\mathbf{P}(\mathbf{R})$ (noting the local field remains finite while $\mathbf{E}\to 0$), we get $\mathbf{E}(\mathbf{R})=0$ — the conductor condition. In practice, taking $\kappa \to \infty$ is a useful check on dielectric calculations and for building intuition. However, caution is needed for situations involving charges embedded in the dielectric bulk (where the screening $\mathbf{E}={\mathbf{E}}_{\text{vac}}/\kappa \to 0$ means the total field vanishes, not just the self-field) or free charge on surfaces (where ${\sigma }_f$ and ${\sigma }_P$ combine in a way that requires care in the limit).

---

The Energy of Dielectric Matter

The total electrostatic energy is important for understanding the thermodynamics of dielectrics and for calculating forces on and within dielectric matter. Two distinct energy concepts arise: the total energy $U_E$ of a polarized dielectric, and the energy change $\Delta U_E$ when a dielectric becomes polarized. The vacuum case (Chapter 3) is recovered by setting $\mathbf{D}={ϵ}_0\mathbf{E}$ everywhere.

Total Energy $U_E$

Derivation I — Charged Conductor in a Dielectric

Embed a charged conductor in an infinite dielectric medium. The work to add charge $\delta Q$ at potential $\phi$ is:

$$\delta U_E=\phi \delta Q={\int }_S\mathrm{d}\mathbf{S}\cdot \delta \mathbf{D}\phi =-{\int }_V{\mathrm{d}}^{3}r\nabla \cdot (\delta \mathbf{D}\phi )$$

(the minus sign is for the volume exterior to the conductor). Since $\nabla \cdot (\delta \mathbf{D})=0$ (no free charge in $V$):

$$\delta U_E={\int }_V{\mathrm{d}}^{3}r\nabla \phi \cdot \delta \mathbf{D}=\int {\mathrm{d}}^{3}r\mathbf{E}\cdot \delta \mathbf{D}$$

where the integral extends over all space (the field vanishes inside the conductor).

Derivation II — Work Against Cohesive Forces

When an external field $\mathbf{E}$ induces polarization current ${\mathbf{j}}_P=\partial \mathbf{P}/\partial t$, the Coulomb force density ${\rho }_P\mathbf{E}$ does work at rate:

$${\rho }_P(\mathbf{r})\mathbf{E}(\mathbf{r})\cdot \mathbf{v}(\mathbf{r})={\mathbf{j}}_P(\mathbf{r})\cdot \mathbf{E}(\mathbf{r})$$

This work is done against the cohesive forces (chemical bonds) that oppose polarization. For an isolated dielectric, the electric field increases the internal energy at rate per unit volume:

$$\frac{\partial u_{\text{mat}}}{\partial t}=\frac{\partial \mathbf{P}}{\partial t}\cdot \mathbf{E}$$

Assuming the field energy density in the dielectric is the same as in vacuum ($u_{\text{field}}=\frac{1}{2}{ϵ}_0{E}^2$), the total energy density change is:

$$\mathrm{d}{u}_E=\mathrm{d}{u}_{\text{field}}+\mathrm{d}{u}_{\text{mat}}={ϵ}_0\mathbf{E}\cdot \mathrm{d}\mathbf{E}+\mathbf{E}\cdot \mathrm{d}\mathbf{P}=\mathbf{E}\cdot \mathrm{d}\mathbf{D}$$

Since the total energy equals the net work to establish the field $\mathbf{D}$:

$$U_E=\int {\mathrm{d}}^{3}r{\int }_0^{\mathbf{D}}\mathbf{E}\cdot \delta \mathbf{D}$$

This cannot be evaluated until we specify a constitutive relation. For a simple dielectric:

$$U_E[\mathbf{D}]=\frac{1}{2}\int {\mathrm{d}}^{3}r\mathbf{E}\cdot \mathbf{D}=\frac{1}{2}\int {\mathrm{d}}^{3}r\frac{|\mathbf{D}{|}^2}{ϵ}$$

Physical Interpretation

Writing the integrand as $|\mathbf{D}{|}^2/ϵ$, we see that the energy is minimized when $ϵ$ is large in the same regions where $|\mathbf{D}|$ is large. This makes physical sense: Coulomb energy is lowered when a charge polarizes the medium to draw opposite charge toward itself (screening we saw with point charge).

Natural Variables of $U_E$

This is analogous to the energy used to compute forces on isolated conductors, which is a natural function of free charges. The same holds here: $U_E$ is a natural function of $\mathbf{D}$, and free charges determine $\mathbf{D}$ via $\nabla \cdot \mathbf{D}={\rho }_f$. Thus $U_E=U_E(\mathbf{D},\mathbf{R})$ or $U_E=U_E(Q_k,\mathbf{R})$, with:

$$\mathbf{E}=\frac{1}{V}\left(\frac{\partial U_E}{\partial \mathbf{D}}\right){|}_{\mathbf{R}},\mathbf{F}=-\left(\frac{\partial U_E}{\partial \mathbf{R}}\right){|}_{\mathbf{D}}\text{or}\mathbf{F}=-\left(\frac{\partial U_E}{\partial \mathbf{R}}\right){|}_Q$$

Application — A Classical Model for Quark Confinement

In quantum chromodynamics (QCD), the vacuum may be modeled as a dielectric medium for color charge with a vanishingly small dielectric constant $\kappa$. Since $\kappa <1$, this medium anti-screens free charge: from $\rho ={\rho }_f/\kappa$, the polarization charge has the same sign as the free charge, enhancing its effect. Coulomb repulsion prevents the charges from overlapping, so a tiny sphere of color charge $q$ (radius $a$) digs itself a vacuum cavity of radius $R$ in the medium.

Free quark model: The fields inside and outside the cavity are:

$${ϵ}_0{E}_{\text{in}}=D_{\text{in}}=\frac{q}{4\pi r^2}=D_{\text{out}}=\kappa {ϵ}_0{E}_{\text{out}}$$

The large $E_{\text{out}}$ when $\kappa \to 0$ arises from the large surface polarization charge. Adding the electrostatic energy and a surface energy $U_S=\gamma \cdot 4\pi R^2$ (with $a\ll R$):

$$U_{\text{tot}}=U_{\text{in}}+U_{\text{out}}+U_S=\frac{q^2}{8\pi {ϵ}_0}\left(\frac{1}{a}+\frac{1}{\kappa R}\right)+\gamma 4\pi R^2$$

Minimizing with respect to $R$ gives a stable minimum at $R^{\ast }\propto {\kappa }^{-1/3}$, with $U_{\text{tot}}(R^{\ast })\propto {\kappa }^{-2/3}$. The divergence of this energy when $\kappa \to 0$ “explains” why free quarks are never seen in isolation.

Meson model (quark-antiquark pair): Two tiny spheres with equal and opposite charge. The energetic preference to eliminate $\mathbf{D}$ from regions with vanishing $\kappa$ is easily accommodated by a cavity containing both charges, with $D_{\text{out}}=0$ when $\kappa =0$. Inside: $\int |\mathbf{E}{|}^2{\mathrm{d}}^{3}r$ has a dipole self-energy (divergent but $R$-independent), vanishing cross terms (by orthogonality of angular integrals), and a uniform-field-like term $\sim p^2/({ϵ}_0{R}^3)$ that wants the cavity to grow. Surface tension $\sim R^2$ opposes this, giving $U_{\text{tot}}=A/R^3+B{R}^2$ with a finite minimum independent of $\kappa$. The total energy of this classical meson is finite as $\kappa \to 0$.

TODO: Blog post about how physics builds upon itself — how surface tension + classical electromagnetism can qualitatively explain quark confinement.

Energy to Polarize: $\Delta U_E$

The energy cost to insert a dielectric sample into a pre-existing field ${\mathbf{E}}_0$ (with ${\mathbf{D}}_0={ϵ}_0{\mathbf{E}}_0$):

Derivation I — Direct Subtraction

$$\Delta U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r(\mathbf{E}\cdot \mathbf{D}-{\mathbf{E}}_0\cdot {\mathbf{D}}_0)$$

Adding and subtracting $\frac{1}{2}(\mathbf{E}\cdot {\mathbf{D}}_0-{\mathbf{E}}_0\cdot \mathbf{D})$: the resulting term $\int {\mathrm{d}}^{3}r\mathbf{E}\cdot (\mathbf{D}-{\mathbf{D}}_0)$ can be shown to equal $\phi (Q-Q_0)$ for the geometry of a conductor embedded in the dielectric. For fixed $Q$ (isolated conductor), this vanishes. The remaining terms, using $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ and ${\mathbf{P}}_0=0$:

$$\Delta U_E=-\frac{1}{2}\int {\mathrm{d}}^{3}r\mathbf{P}\cdot {\mathbf{E}}_0$$

Derivation II — Physical Decomposition

For a small linear dielectric ($\mathbf{p}=\alpha {ϵ}_0{\mathbf{E}}_0$), the energy has three contributions:

1. Interaction with external field: $U_{\text{ext}}=-\mathbf{p}\cdot {\mathbf{E}}_0=-p{E}_0$ (the energy gain from placing the dipole in the field).
2. Internal cohesive cost: $U_{\text{int}}=\zeta p^2$ (must be at least quadratic to produce a stable minimum; $\zeta >0$).

Minimizing $U_E=-p{E}_0+\zeta p^2$ with respect to $p$ and using $p=\alpha {ϵ}_0{E}_0$:

$$U_E=-\frac{1}{2}\mathbf{p}\cdot {\mathbf{E}}_0=-\frac{1}{2}\alpha {ϵ}_0|{\mathbf{E}}_0{|}^2$$

The factor of $\frac{1}{2}$ (compared to $-\mathbf{p}\cdot {\mathbf{E}}_0$) arises because half the interaction energy goes into paying the internal cost of creating the dipole.

For a finite-sized sample, there is also the self-field of the dielectric. With $\mathbf{E}={\mathbf{E}}_0+{\mathbf{E}}_{\text{ind}}$:

• $U_{\text{ext}}=-\int {\mathrm{d}}^{3}r\mathbf{P}(\mathbf{r})\cdot {\mathbf{E}}_0(\mathbf{r})$
• $U_{\text{self}}=-\frac{1}{2}\int {\mathrm{d}}^{3}r\mathbf{P}(\mathbf{r})\cdot {\mathbf{E}}_{\text{ind}}(\mathbf{r})$ (the $\frac{1}{2}$ avoids double-counting dipole-dipole interactions)
• $U_{\text{int}}=\int {\mathrm{d}}^{3}r{\int }_0^{\mathbf{P}}\delta \mathbf{P}\cdot \mathbf{E}=\frac{1}{2}\int {\mathrm{d}}^{3}r\mathbf{P}\cdot ({\mathbf{E}}_0+{\mathbf{E}}_{\text{ind}})$ (by linearity, $\mathbf{E}$ ramps linearly with $\mathbf{P}$)

The self-energy cancels between $U_{\text{self}}$ and $U_{\text{int}}$, and $U_{\text{int}}$ contributes $+\frac{1}{2}\int \mathbf{P}\cdot {\mathbf{E}}_0$ which halves $U_{\text{ext}}$, giving the same result.

(Why is the self-field irrelevant for a point dipole? A point dipole has no spatial extent, so there is no region where its own field acts on its own charge distribution — the self-field energy is a constant that doesn’t depend on ${\mathbf{E}}_0$. TODO: Is this really sensible?)

Energy at Fixed Potential

For $N$ conductors held at fixed potentials ${\phi }_k$, the Legendre transform gives:

$${\hat{U}}_E=U_E-\int {\mathrm{d}}^{3}r\mathbf{D}\cdot \mathbf{E}=-\frac{1}{2}\int {\mathrm{d}}^{3}r\mathbf{D}\cdot \mathbf{E}$$

Varying: $\delta {\hat{U}}_E=-\int {\mathrm{d}}^{3}r\mathbf{D}\cdot \delta \mathbf{E}$, so ${\hat{U}}_E$ is a natural function of $\mathbf{E}$ (or equivalently ${\phi }_k$), with:

$$\mathbf{D}=-\frac{1}{V}\left(\frac{\partial {\hat{U}}_E}{\partial \mathbf{E}}\right){|}_{\mathbf{R}},\mathbf{F}=-\left(\frac{\partial {\hat{U}}_E}{\partial \mathbf{R}}\right){|}_{\mathbf{D}}\text{or}\mathbf{F}=+\left(\frac{\partial {\hat{U}}_E}{\partial \mathbf{R}}\right){|}_{\phi }$$

Note the sign flip: at fixed potential, $\mathbf{F}=+\partial {\hat{U}}_E/\partial \mathbf{R}$, as in the conductor case (Chapter 5).

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Forces on Dielectric Matter

Punchline: The net force on an isolated dielectric sample is fairly straightforward to calculate, but subtleties arise for forces on a sub-volume of a dielectric. We treat these cases separately.

Force on an Isolated Sample

For a sample with polarization $\mathbf{P}(\mathbf{r})$ in an external field ${\mathbf{E}}_0(\mathbf{r})$ (whether $\mathbf{P}$ is induced by ${\mathbf{E}}_0$ or not), the Coulomb force is:

$$\mathbf{F}={\int }_V{\mathrm{d}}^{3}r({\rho }_f-\nabla \cdot \mathbf{P}){\mathbf{E}}_0+{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{P}({\mathbf{r}}_S){\mathbf{E}}_0({\mathbf{r}}_S)$$

or equivalently, using the dipole-distribution viewpoint:

$$\mathbf{F}={\int }_V{\mathrm{d}}^{3}r[{\rho }_f(\mathbf{r})+(\mathbf{P}(\mathbf{r})\cdot \nabla )]{\mathbf{E}}_0(\mathbf{r})$$

(These are equivalent by ${\int }_S\mathrm{d}S(\hat{\mathbf{n}}\cdot \mathbf{a})\mathbf{b}={\int }_V{\mathrm{d}}^{3}r[(\nabla \cdot \mathbf{a})\mathbf{b}+(\mathbf{a}\cdot \nabla )\mathbf{b}]$.)

Since the dielectric cannot exert a net force on itself, we can replace ${\mathbf{E}}_0$ by the full field $\mathbf{E}$ in the volume integral. In the surface integral, the correct replacement is ${\mathbf{E}}_0\to {\mathbf{E}}_{\text{avg}}=\frac{1}{2}({\mathbf{E}}_{\text{in}}+{\mathbf{E}}_{\text{out}})$, since the average of the self-field on the surface equals the average of its inner and outer limits:

$$\mathbf{F}={\int }_V{\mathrm{d}}^{3}r({\rho }_f-\nabla \cdot \mathbf{P})\mathbf{E}(\mathbf{r})+{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{P}({\mathbf{r}}_S){\mathbf{E}}_{\text{avg}}({\mathbf{r}}_S)$$

Integrating the volume term by parts gives ${\int }_V{\mathrm{d}}^{3}r[{\rho }_f+(\mathbf{P}\cdot \nabla )]\mathbf{E}$, and the remaining surface terms combine (using the fact that $\mathbf{E}$ in the volume integral is ${\mathbf{E}}_{\text{in}}$ evaluated at $S$) into:

$${\int }_S\mathrm{d}S\cdot \mathbf{P}\cdot \frac{1}{2}({\mathbf{E}}_{\text{out}}-{\mathbf{E}}_{\text{in}})=\frac{1}{2{ϵ}_0}{\int }_S\mathrm{d}S[\hat{\mathbf{n}}({\mathbf{r}}_S)\cdot \mathbf{P}({\mathbf{r}}_S){]}^2$$

where we used the discontinuity condition ${\mathbf{E}}_{\text{out}}-{\mathbf{E}}_{\text{in}}=(\hat{\mathbf{n}}\cdot \mathbf{P})\hat{\mathbf{n}}/{ϵ}_0={\sigma }_P\hat{\mathbf{n}}/{ϵ}_0$.

Validity of Bilinear Force Expressions

These bilinear (product of two macroscopic fields) force expressions raise macroscopic validity questions — they must be treated as independent assumptions subject to experimental verification, since at a microscopic level the actual Coulomb force involves the true microscopic fields, and the passage to macroscopic averages in products of fields is not guaranteed to commute with the averaging.

Example — Force Between a Charged Object and an Uncharged Dielectric

At large distance $r$ (compared to both object sizes), ${\mathbf{E}}_0\approx (Q/4\pi {ϵ}_0)\hat{\mathbf{r}}/r^2$ and:

$$\mathbf{F}\approx (\mathbf{p}\cdot \nabla ){\mathbf{E}}_0$$

For a linear dielectric, $\mathbf{p}=\gamma {\mathbf{E}}_0$ ($\gamma$ constant), and using $\nabla \times {\mathbf{E}}_0=0$:

$$\mathbf{F}=\frac{1}{2}\gamma \nabla |{\mathbf{E}}_0{|}^2\propto -\frac{Q^2\hat{\mathbf{r}}}{r^5}$$

This force is always attractive, independent of the sign of $Q$: the charged object attracts unlike polarization charge (which is nearer) more strongly than it repels like polarization charge.

With other choices of non-uniform ${\mathbf{E}}_0(\mathbf{r})$, the force $\mathbf{F}=\frac{1}{2}\gamma \nabla |{\mathbf{E}}_0{|}^2$ has been used to sort, position, and transport minute dielectric bodies, including living cells. This phenomenon is known as dielectrophoresis (from the Greek for “to bear”).

TODO: I forgot to paste my interpretations/derivations of many of these examples so they are too similar to text. Rewrite.

Why the Isolated-Sample Formula Fails for Sub-Volumes

Don't Use the Coulomb Formula on a Sub-Volume

The expressions $\mathbf{F}=\int {\mathrm{d}}^{3}r({\rho }_f+{\rho }_P)\mathbf{E}$ and $\mathbf{F}=\int {\mathrm{d}}^{3}r{\rho }_f\mathbf{E}+(\mathbf{P}\cdot \nabla )\mathbf{E}$ are correct only when integrated over an isolated, complete sample. Restricting them to a sub-volume of a larger dielectric omits short-range cohesive forces at the cut surface and gives wrong answers (often off by a factor of $\kappa$ or 2). For sub-volumes use the Helmholtz force or the Maxwell stress tensor instead.

Applying the isolated-sample formula to find the force per unit area at a ${\kappa }_1/{\kappa }_2$ interface (e.g., inside a parallel-plate capacitor with ${\sigma }_f$ on the plates), only the surface term survives near the boundary, giving $f={\sigma }_P{E}_0$. Using $E_0={\sigma }_f/{ϵ}_0$:

$$f=\frac{{\sigma }_f^2}{{ϵ}_0}\left(\frac{1}{{\kappa }_2}-\frac{1}{{\kappa }_1}\right)$$

In the limit ${\kappa }_1\to \infty$ (one side conducting): $f\to {\sigma }_f^2/({\kappa }_2{ϵ}_0)$. But the correct result (from energy methods with $C={\kappa }_2{ϵ}_{0}A/d$ and $\mathbf{R}=-\hat{\mathbf{z}}d$, giving $F=Q^2/(2C)\partial C/\partial R$) is:

$$f=\frac{{\sigma }_f^2}{2{\kappa }_2{ϵ}_0}\text{(off by a factor of 2!)}$$

Physical reason: When applied to a sub-volume, the Coulomb force formulae omit the effect of short-range, non-electrostatic forces (quantum-mechanical cohesive forces) that act on the sub-volume boundary. These forces are polarization-dependent and thus field-dependent. They cancel exactly when summed over all sub-volumes (giving the correct net force on the entire body), but contribute to the force on any individual sub-volume.

The Helmholtz Force

To properly account for the sub-volume forces, we use a variational approach: virtual displacement $\delta \mathbf{r}$ of the dielectric gives $\delta U_E=-\mathbf{F}\cdot \delta \mathbf{r}$.

Since $U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r|\mathbf{D}{|}^2/ϵ$:

$$\delta U_E=\int {\mathrm{d}}^{3}r\frac{\mathbf{D}\cdot \delta \mathbf{D}}{ϵ}-\frac{1}{2}\int {\mathrm{d}}^{3}r\frac{|\mathbf{D}{|}^2}{{ϵ}^2}\delta ϵ=\int {\mathrm{d}}^{3}r\mathbf{E}\cdot \delta \mathbf{D}-\frac{1}{2}\int {\mathrm{d}}^{3}r|\mathbf{E}{|}^2\delta ϵ$$

For variations induced by virtual displacement: $\delta \mathbf{D}=-(\delta \mathbf{r}\cdot \nabla )\mathbf{D}$ and $\delta ϵ=-(\delta \mathbf{r}\cdot \nabla )ϵ$. Using $\nabla \times (\delta \mathbf{r}\times \mathbf{D})=\delta \mathbf{r}(\nabla \cdot \mathbf{D})-(\delta \mathbf{r}\cdot \nabla )\mathbf{D}$, we arrive at the Helmholtz formula:

$$\mathbf{F}=\int {\mathrm{d}}^{3}r{\rho }_f(\mathbf{r})\mathbf{E}(\mathbf{r})-\frac{1}{2}\int {\mathrm{d}}^{3}r|\mathbf{E}(\mathbf{r}){|}^2\nabla ϵ(\mathbf{r})$$

When the domain of integration is restricted to a sub-volume, this formula gives the mechanical force on that volume. The field exerts a force only on free charge and at points where $ϵ$ varies in space — the $\nabla ϵ$ term encodes the contribution from abrupt dielectric interfaces and implicitly includes the non-electrostatic cohesive forces.

Why the Energy Variation Captures Internal Forces

The total energy $U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r|\mathbf{D}{|}^2/ϵ$ is integrated over all space. When we vary $ϵ$ by displacing a sub-volume, the energy change accounts for field rearrangements everywhere — including the work done by and against the short-range forces at the sub-volume boundary. These forces maintain mechanical equilibrium of the dielectric and are implicitly contained in the constitutive relation $\mathbf{D}=ϵ\mathbf{E}$. The Coulomb force formulae, by contrast, only sum explicit electromagnetic forces on charges, missing the field-dependent part of the cohesive forces.

Example — Force at a Dielectric Interface (Corrected)

Returning to the ${ϵ}_1/{ϵ}_2$ interface at $z=0$ in a parallel-plate capacitor, the Helmholtz force per unit area is $f=f\hat{\mathbf{z}}$ where:

$$f=-\frac{1}{2}{\int }_{0^{-}}^{0^{+}}{E}_z^2\frac{\mathrm{d}ϵ}{\mathrm{d}z}\mathrm{d}z$$

Since $E_z$ is discontinuous at $z=0$ but $D_z={\sigma }_f$ is continuous (Gauss’s law), rewrite in terms of $\mathbf{D}$:

$$f=-\frac{1}{2}{\int }_{0^{-}}^{0^{+}}\frac{D_z^2}{{ϵ}^2}\frac{\mathrm{d}ϵ}{\mathrm{d}z}\mathrm{d}z=\frac{1}{2}{\int }_{0^{-}}^{0^{+}}{D}_z^2\frac{\mathrm{d}}{\mathrm{d}z}\frac{1}{ϵ}\mathrm{d}z=\frac{{\sigma }_f^2}{2{ϵ}_0}\left(\frac{1}{{\kappa }_2}-\frac{1}{{\kappa }_1}\right)$$

Independent of the field direction, this force tends to increase the volume of the medium with large $ϵ$ and decrease the volume of the medium with small $ϵ$.

In the ${\kappa }_1\to \infty$ limit, with $E_0={\sigma }_f/{ϵ}_0$:

$$f\to \frac{1}{2}\frac{E_0}{{\kappa }_2}{\sigma }_f$$

matching the expected force of attraction per unit area between capacitor plates filled with dielectric constant ${\kappa }_2$.

The Maxwell Stress Tensor for Dielectrics

The Helmholtz formula contains $\nabla ϵ$, which requires knowing (and integrating over) where $ϵ$ varies — inconvenient for complex geometries. To eliminate spatial derivatives, use:

$$\frac{E^2}{2}{\partial }_jϵ=\frac{1}{2}{\partial }_j(D_k{E}_k)-D_k{\partial }_j{E}_k$$

and the curl trick: $\nabla \times \mathbf{E}=0$ implies ${\partial }_j{E}_k={\partial }_k{E}_j$, so $(D_k{\partial }_j)E_k=(D_k{\partial }_k)E_j=(\mathbf{D}\cdot \nabla )E_j$. This gives:

$$\mathbf{F}=\int {\mathrm{d}}^{3}r(\nabla \cdot \mathbf{D})\mathbf{E}-\int {\mathrm{d}}^{3}r(\mathbf{D}\cdot \nabla )\mathbf{E}-\frac{1}{2}\int {\mathrm{d}}^{3}r\nabla (\mathbf{D}\cdot \mathbf{E})$$

Reducing to surface integrals:

$$\mathbf{F}={\int }_S\mathrm{d}S\left[(\hat{\mathbf{n}}\cdot \mathbf{D})\mathbf{E}-\frac{1}{2}\hat{\mathbf{n}}(\mathbf{D}\cdot \mathbf{E})\right]={\int }_S\mathrm{d}S\hat{\mathbf{n}}\cdot \mathbf{T}(\mathbf{D})$$

where the electric stress tensor for dielectric matter generalizes the vacuum one:

$$T_{ij}(\mathbf{D}):=D_i{E}_j-\frac{1}{2}{\delta }_{ij}\mathbf{D}\cdot \mathbf{E}$$

This is symmetric for simple media ($\mathbf{D}=ϵ\mathbf{E}$). Compared to the Helmholtz formula, it contains no spatial derivatives and requires only a surface integral — making it simple to evaluate with clever choices of $S$.

Example — Force on an Embedded Dielectric (Stress Tensor)

A point charge $q$ is embedded in a dielectric with permittivity $ϵ=\kappa {ϵ}_0$. A second dielectric (${ϵ}^{\prime }={\kappa }^{\prime }{ϵ}_0$) containing a charge $q^{\prime }$ is embedded inside the first. To find the force on the embedded dielectric, choose $S$ (dashed) infinitesimally larger than its surface. Since $\mathbf{D}=ϵ\mathbf{E}$ everywhere on $S$, we can pull $ϵ$ outside the integral and convert to a volume integral:

$$F_j=ϵ{\int }_{\Omega }{\mathrm{d}}^{3}r{\partial }_i(E_i{E}_j-\frac{1}{2}{\delta }_{ij}{E}^2)$$

Reversing the algebraic steps from Section 3.7:

$$\mathbf{F}=\kappa {\int }_{\Omega }{\mathrm{d}}^{3}r\rho (\mathbf{r})\mathbf{E}(\mathbf{r})$$

where $\rho ={\rho }_f+{\rho }_P$ is the total charge in $\Omega$. This resembles the Coulomb force but with an extra factor of $\kappa$ — this factor (non-obviously) accounts for the field-dependent short-range force exerted by the $\kappa$ medium on the ${\kappa }^{\prime }$ medium.

Summary of Force Formulae

Formula	Valid for	Method
$\mathbf{F}={\int }_V{\mathrm{d}}^{3}r({\rho }_f+\mathbf{P}\cdot \nabla ){\mathbf{E}}_0$	Isolated sample only	Direct Coulomb (external field)
$\mathbf{F}={\int }_V({\rho }_f-\nabla \cdot \mathbf{P})\mathbf{E}{\mathrm{d}}^{3}r+{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{P}{\mathbf{E}}_{\text{avg}}$	Isolated sample only	Coulomb (total field)
$\mathbf{F}=\int {\mathrm{d}}^{3}r{\rho }_f\mathbf{E}-\frac{1}{2}\int {\mathrm{d}}^{3}r{\mathbf{E}}^2\nabla ϵ$	Any volume (including sub-volumes)	Helmholtz (energy variation)
$\mathbf{F}={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{T}(\mathbf{D})$	Any volume (including sub-volumes)	Stress tensor

The first two formulae give wrong results for sub-volumes because they miss field-dependent non-electrostatic (cohesive) forces at internal boundaries. The Helmholtz formula and stress tensor approach correctly include these effects.

Example — Dielectric Slab Pulled into a Capacitor

A slab of simple dielectric is partially inserted (distance $s$) into the gap $d$ between the square ($L\times L$) plates of a capacitor at fixed potential $V$. Ignoring fringe fields:

Method I (Polarization energy): With $|\mathbf{E}|=|{\mathbf{E}}_0|=V/d$ inside the capacitor:

$${\hat{U}}_E=-\frac{1}{2}{\int }_V{\mathrm{d}}^{3}r\mathbf{P}\cdot {\mathbf{E}}_0=-\frac{1}{2}{ϵ}_0(\kappa -1)\frac{V^2}{d^2}\times (Lds)$$

So $F=-\partial {\hat{U}}_E/\partial s={ϵ}_0(\kappa -1)L{V}^2/(2d)$, directed into the capacitor.

Method II (Helmholtz): No free charge in the dielectric, $\mathbf{E}=(V/d)\hat{\mathbf{z}}$ inside the capacitor. Write $ϵ(x)={ϵ}_0\Theta (x-s)+ϵ\Theta (s-x)$, so $\nabla ϵ={ϵ}_0(1-\kappa )\delta (x-s)\hat{\mathbf{x}}$. Since $\mathbf{E}$ is continuous at $x=s$ (no free surface charge there, and the field is tangential):

$$F=-\frac{1}{2}Ld{\left(\frac{V}{d}\right)}^2{ϵ}_0(1-\kappa )={ϵ}_0(\kappa -1)\frac{L{V}^2}{2d}$$

Physical origin of the force: Neither method clearly identifies it. Using the fundamental force law $\mathbf{F}={\int }_V(\mathbf{P}\cdot \nabla ){\mathbf{E}}_0{\mathrm{d}}^{3}r$, we see that the force can only come from regions where the capacitor field deviates from uniformity — i.e., the force arises entirely from the fringing field near the plate edges. The uniform field inside cannot exert a net force on the dipole distribution; it is the field gradient at the boundary of the capacitor that pulls the dielectric in.

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Problems

Polarization and Bound Charge

6.1 — Polarization by Superposition

Two spheres of radius $R$ have uniform but equal and opposite charge densities $\pm \rho$. Their centers are displaced by an infinitesimal vector $\mathbf{\delta }$. Show that the resulting electric field is identical to that of a sphere with uniform polarization $\mathbf{P}$.

Solution Sketch

The total charge density is $\rho (\mathbf{r})-\rho (\mathbf{r}-\mathbf{\delta })$. Taylor-expanding to first order in $\mathbf{\delta }$:

$$\rho (\mathbf{r})-\rho (\mathbf{r}-\mathbf{\delta })\approx (\mathbf{\delta }\cdot \nabla )\rho (\mathbf{r})$$

The Gauss’s law electric field of a uniformly charged sphere (density $\rho$, centered at the origin) is:

$$\mathbf{E}(\mathbf{r})=\{\begin{array}{ll}\frac{\rho \mathbf{r}}{3{ϵ}_0}& r<R\\ \frac{\rho R^3}{3{ϵ}_0}\frac{\hat{\mathbf{r}}}{r^2}& r>R\end{array}$$

Superposing the fields of $+\rho$ centered at the origin and $-\rho$ centered at $\mathbf{\delta }$, and expanding $|\mathbf{r}-\mathbf{\delta }{|}^{-3}\approx r^{-3}(1+3\hat{\mathbf{r}}\cdot \mathbf{\delta }/r+\cdots )$ for the exterior:

• Inside: $\mathbf{E}=\rho \mathbf{\delta }/(3{ϵ}_0)$
• Outside: dipolar field with moment $\mathbf{p}=\rho V\mathbf{\delta }$

Identifying $\mathbf{P}=-\rho \mathbf{\delta }$ (so that ${\mathbf{E}}_{\text{in}}=-\mathbf{P}/(3{ϵ}_0)$), this matches exactly the field of a uniformly polarized sphere with polarization $\mathbf{P}$.

6.3 — The Energy of a Polarized Ball

Find the total electrostatic energy of a ball with radius $R$ and uniform polarization $\mathbf{P}$.

Solution Sketch

Choose $\mathbf{P}=P\hat{\mathbf{z}}$. The only charge is the surface polarization charge density ${\sigma }_P(\theta )=\mathbf{P}\cdot \hat{\mathbf{r}}=P\cos \theta$. The volume polarization charge vanishes. The total energy is:

$$U_E=\frac{1}{2}{\int }_S\mathrm{d}S\sigma (\theta )\phi ({\mathbf{r}}_S)=\frac{1}{2}\frac{1}{4\pi {ϵ}_0}{\oint }_S\mathrm{d}S{\int }_{S^{\prime }}\mathrm{d}{S}^{\prime }\frac{\sigma (\theta )\sigma ({\theta }^{\prime })}{|{\mathbf{r}}_S-{\mathbf{r}}_S^{\prime }|}$$

Since both $r_S=r_S^{\prime }=R$, expand the Green function in spherical harmonics:

$$\frac{1}{|{\mathbf{r}}_S-{\mathbf{r}}_S^{\prime }|}=\sum _{L=0}^{\infty }\sum _{M=-L}^L\frac{4\pi }{2L+1}\frac{1}{R}{Y}_{LM}^{\ast }({\theta }^{\prime },{\varphi }^{\prime })Y_{LM}(\theta ,\varphi )$$

Since $\cos \theta =\sqrt{4\pi /3}{Y}_{10}(\theta ,\varphi )$, both surface integrals pick out only $L=1$, $M=0$ by orthonormality of the spherical harmonics. The result is:

$$U_E=\frac{P^2{R}^3}{9{ϵ}_0}$$

6.4 — A Hole in Radially Polarized Matter

The polarization in all of space has the form $\mathbf{P}=P\Theta (r-R)\hat{\mathbf{r}}$, where $P$ and $R$ are constants. Find the polarization charge density and the electric field everywhere.

Solution Sketch

Charges: $\nabla \cdot \mathbf{P}=P[\delta (r-R)+2\Theta (r-R)/r]$, giving ${\rho }_P=-2P/r$ for $r>R$ and ${\sigma }_P=P$ at $r=R$.

Field by Gauss’s law: Total enclosed charge for $r>R$: $Q(r)=4\pi R^{2}P-4\pi P(r^2-R^2)=4\pi P(2R^2-r^2)$.

$$\mathbf{E}(\mathbf{r})=\{\begin{array}{ll}0& r<R\\ \frac{P}{{ϵ}_0}\left(\frac{2R^2}{r^2}-1\right)\hat{\mathbf{r}}& r>R\end{array}$$

Physical insight: The field vanishes inside the hole — the surface charge exactly cancels the volume charge contribution, creating a perfect shield despite the nontrivial exterior charge distribution.

6.5 — The Field at the Center of a Polarized Cube

A cube is uniformly polarized parallel to one of its edges. Show that $\mathbf{E}(0)=-\mathbf{P}/(3{ϵ}_0)$. Compare with a uniformly polarized sphere.

Solution Sketch

Only ${\sigma }_P=\pm P$ on the two faces $\perp \hat{\mathbf{z}}$ contributes ($\nabla \cdot \mathbf{P}=0$ in the bulk, ${\sigma }_P=0$ on the other four faces).

Solid angle trick: The $z$-component of the field from one charged face is $E_z\propto \int \mathrm{d}\Omega$, where $\mathrm{d}\Omega =\mathrm{d}{S}^{\prime }\cos \alpha /r^{\prime 2}$ is the solid angle subtended at the center. By cubic symmetry, each face subtends $4\pi /6$. Both charged faces push $\mathbf{E}$ in the $-\hat{\mathbf{z}}$ direction:

$$E_z(0)=-\frac{2P}{4\pi {ϵ}_0}\cdot \frac{4\pi }{6}=-\frac{P}{3{ϵ}_0}$$

Identical to the uniformly polarized sphere — any shape with cubic or higher symmetry gives the same interior field at its center.

Fields and Potentials

6.7 — Isotropic Polarization

The polarization inside an origin-centered sphere of radius $R$ is $\mathbf{P}(\mathbf{r})=P(r)\hat{\mathbf{r}}$.

• (a) Show that $\phi (\mathbf{r})$ outside the sphere equals the potential of a point dipole at the origin with moment $\mathbf{p}={\int }_V{\mathrm{d}}^{3}r\mathbf{P}$.
• (b) Find $\phi (\mathbf{r})$ inside the sphere.

Solution Sketch

Key tool: The angular integral $\int \mathrm{d}{\Omega }^{\prime }/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ is the potential of a uniform shell (radius $r^{\prime }$, unit charge): $4\pi /r$ for $r>r^{\prime }$ and $4\pi /r^{\prime }$ for $r<r^{\prime }$.

(a) For $r>R$: all source shells satisfy $r>r^{\prime }$, so the angular integral pulls out $1/r$. The radial integral then yields $\mathbf{p}={\int }_V{\mathrm{d}}^3{r}^{\prime }\mathbf{P}$, giving:

$$\phi (\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{\mathbf{p}\cdot \hat{\mathbf{r}}}{r^2}$$

(b) For $r<R$: shells with $r^{\prime }>r$ contribute a potential independent of $\mathbf{r}$ (constant); shells with $r^{\prime }<r$ contribute zero (the angular average $\int \mathrm{d}{\Omega }^{\prime }{\hat{\mathbf{r}}}^{\prime }=0$). So $\phi =0$ inside.

Physical insight: Each shell’s contribution is isotropic when viewed from inside — the angular average of ${\hat{\mathbf{r}}}^{\prime }$ vanishes, producing zero interior potential.

6.8 — $\mathbf{E}$ and $\mathbf{D}$ for an Annular Dielectric

(a) The volume between concentric shells (radii $R_1<R_2$) has uniform polarization $\mathbf{P}$. Find $\mathbf{E}$ everywhere. (b) The volume inside a sphere of radius $R$ has $\mathbf{P}=P\hat{\mathbf{r}}/r^2$. Find $\mathbf{D}$ everywhere.

Solution Sketch

(a) Uniform $\mathbf{P}$ means ${\rho }_P=-\nabla \cdot \mathbf{P}=0$ in the bulk. Surface charges ${\sigma }_P=\mathbf{P}\cdot \hat{\mathbf{n}}$ appear on both shells. Each surface carries a $\cos \theta$ charge density, producing a uniform interior field and a dipole exterior field (as in Application 6.1). The total field is the superposition from both shells — the inner shell’s field contributes inside $R_1$ and outside $R_1$, the outer shell’s inside $R_2$ and outside $R_2$, etc.

(b) For $\mathbf{P}=P\hat{\mathbf{r}}/r^2$: the free charge is zero, so $\nabla \cdot \mathbf{D}=0$. Moreover:

$$\nabla \times \mathbf{D}=\nabla \times \mathbf{P}=\nabla \times \left(\frac{P}{r^2}\hat{\mathbf{r}}\right)=0$$

since $P\hat{\mathbf{r}}/r^2\propto \nabla (1/r)$, which is curl-free. With both divergence and curl vanishing and appropriate boundary conditions at infinity, by the Helmholtz theorem:

$$\mathbf{D}=0\text{everywhere}$$

Key insight: A polarization that is itself a gradient field ($\nabla \times \mathbf{P}=0$) with no free charge gives $\mathbf{D}=0$ — the displacement field need not be nonzero even when $\mathbf{E}$ and $\mathbf{P}$ are individually nonzero.

Measurement and Matching Conditions

6.9 — The Correct Way to Define $\mathbf{E}$

A charge $q$ placed at a point measures a force ${\mathbf{F}}_q$, and $-q$ at the same point measures ${\mathbf{F}}_{-q}$. Nearby conductors or dielectrics are present. Derive an expression for $\mathbf{E}$ in terms of ${\mathbf{F}}_q$ and ${\mathbf{F}}_{-q}$, without requiring $q\to 0$.

Solution Sketch

In the presence of charge $q$, nearby matter polarizes and creates an induced field ${\mathbf{E}}_{\text{ind}}$ at the position of $q$. The measured force is:

$${\mathbf{F}}_q=q(\mathbf{E}+{\mathbf{E}}_{\text{ind}})$$

The linearity of electrostatics guarantees that the entire chain — $q$ produces a field, the field determines induced charges, induced charges produce ${\mathbf{E}}_{\text{ind}}$ — is linear in $q$. So when $q\to -q$, ${\mathbf{E}}_{\text{ind}}$ changes sign:

$${\mathbf{F}}_{-q}=-q(\mathbf{E}-{\mathbf{E}}_{\text{ind}})$$

Adding the two equations cancels ${\mathbf{E}}_{\text{ind}}$:

$$\mathbf{E}=\frac{{\mathbf{F}}_q-{\mathbf{F}}_{-q}}{2q}$$

Physical insight: The conventional definition $\mathbf{E}={\lim }_{q\to 0}{\mathbf{F}}_q/q$ is operationally impossible — you can’t measure a force on a vanishing charge. This two-measurement protocol exploits linearity to subtract out the induced field exactly, at finite $q$.

TODO: Turn this into a blog post — it’s a beautiful example of how linearity enables clever experimental design.

Source: W.M. Saslow, Electricity, Magnetism, and Light (Academic, Amsterdam, 2002).

6.11 — Cavity Field

A uniform field ${\mathbf{E}}_0$ exists throughout a homogeneous dielectric with permittivity $ϵ$. Find the field inside a vacuum cavity shaped as a thin rectangular pancake ($L\times L\times h$, $h\ll L$). Express ${\mathbf{E}}_{\text{cav}}$ in terms of ${\mathbf{E}}_0$, $\hat{\mathbf{n}}$, $ϵ$, and ${ϵ}_0$.

Solution Sketch

When $h\ll L$, only the matching conditions at the large flat faces (normal $\hat{\mathbf{n}}$) matter. Decompose the field into normal and tangential parts:

$$\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }=(\mathbf{E}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}+[\mathbf{E}-(\mathbf{E}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}]$$

The matching conditions are: $E_{\parallel }$ is continuous, and $D_{\perp }=[ϵE{]}_{\perp }$ is continuous. Thus:

$${\mathbf{E}}_{\text{cav}}=\frac{ϵ}{{ϵ}_0}({\mathbf{E}}_0\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}+[{\mathbf{E}}_0-({\mathbf{E}}_0\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}]$$

Physical picture: The pancake cavity acts like two parallel sheets of bound surface charge $\pm {\sigma }_P$. These sheets produce a field $\sim {\sigma }_P/{ϵ}_0$ that supplements ${\mathbf{E}}_0$ in the normal direction, boosting it by the factor $ϵ/{ϵ}_0=\kappa$. The tangential component, unaffected by the thin gap, passes through unchanged.

Boundary Value Problems

6.14 — Spherical Conductor Embedded in a Dielectric

A spherical conductor of radius $R_1$ is surrounded by a dielectric ($\kappa$) extending to $R_2$.

• (a) The conductor has charge $Q$. Find $\mathbf{E}$ everywhere and confirm zero total polarization charge.
• (b) The conductor is grounded and placed in a uniform field ${\mathbf{E}}_0$. Find the potential everywhere and the charge drawn from ground.

Solution Sketch

(a) Gauss’s law ($\int \mathrm{d}\mathbf{S}\cdot \mathbf{D}=Q_{f,\text{enc}}$) with spherical symmetry and $\mathbf{D}=\kappa {ϵ}_0\mathbf{E}$:

$$\mathbf{E}(\mathbf{r})=\frac{Q\hat{\mathbf{r}}}{4\pi r^2}\times \{\begin{array}{ll}0& r<R_1\\ 1/(\kappa {ϵ}_0)& R_1<r<R_2\\ 1/{ϵ}_0& r>R_2\end{array}$$

The bulk polarization charge is ${\rho }_P=-(\kappa -1)\nabla \cdot \mathbf{E}=0$ everywhere (no free charge in the dielectric). Surface polarization charges:

$${\sigma }_P(R_1)=-(\kappa -1)E(R_1)=-\frac{\kappa -1}{\kappa }\frac{Q}{4\pi {ϵ}_0{R}_1^2}$$ $${\sigma }_P(R_2)=+(\kappa -1)E(R_2)=+\frac{\kappa -1}{\kappa }\frac{Q}{4\pi {ϵ}_0{R}_2^2}$$

These integrate to $-(\kappa -1)Q/\kappa$ and $+(\kappa -1)Q/\kappa$ respectively — the dielectric is neutral.

(b) Azimuthal symmetry from ${\mathbf{E}}_0$ forces $\ell =1$ Legendre structure. Ansatz: ${\phi }_{\text{in}}=B(r-R_1^3/r^2)\cos \theta$ (vanishes at $r=R_1$), ${\phi }_{\text{out}}=(-E_{0}r+A/r^2)\cos \theta$. Matching $\phi$ and $\kappa \partial \phi /\partial r$ at $R_2$ gives $A$ and $B$. The induced conductor charge ${\sigma }_f\propto \cos \theta$ integrates to zero — no net charge drawn from ground.

Physical insight: The response is purely dipolar; grounding provides a path for charge flow, but the monopole moment vanishes by symmetry — there is no energetic reason to accumulate net charge.

6.15 — Parallel-Plate Capacitor with an Air Gap

An air-gap capacitor (plate area $A$, separation $d$) breaks down at voltage $V_0$. A dielectric slab ($\kappa$, thickness $t<d$) is inserted. Find the new breakdown voltage.

Solution Sketch

The critical field for breakdown is $E_{\text{crit}}=V_0/d$ (the field in the air gap at breakdown without the dielectric).

With the slab inserted, the field in the air gap is $E_{\text{air}}$ and in the dielectric is $E_d=E_{\text{air}}/\kappa$ (from continuity of $D_{\perp }$ and absence of free charge at the interface). The voltage across the capacitor is:

$$V=E_{\text{air}}(d-t)+\frac{E_{\text{air}}}{\kappa }t=E_{\text{air}}\left(d-t+\frac{t}{\kappa }\right)$$

Breakdown occurs when $E_{\text{air}}=E_{\text{crit}}=V_0/d$:

$$V_{\text{breakdown}}=\frac{V_0}{d}\left(d-t+\frac{t}{\kappa }\right)=V_0\left(1-\frac{t}{d}\left(1-\frac{1}{\kappa }\right)\right)$$

Since $\kappa >1$, the new breakdown voltage is lower than $V_0$. The dielectric concentrates more of the voltage drop across the (thinner) air gap — a cautionary result for high-voltage design.

Helmholtz Theorem and Formal Structure

6.16 — Helmholtz Theorem for $\mathbf{D}(\mathbf{r})$

Write the Helmholtz decomposition for $\mathbf{D}(\mathbf{r})$ and simplify for simple dielectric matter.

Solution Sketch

From $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{D}=\nabla \times \mathbf{P}$, the Helmholtz theorem gives:

$$\mathbf{D}(\mathbf{r})=-\nabla \int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_f({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}+\nabla \times \int {\mathrm{d}}^3{r}^{\prime }\frac{{\nabla }^{\prime }\times \mathbf{P}({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

For simple dielectric matter, $\mathbf{P}={ϵ}_0\chi \mathbf{E}$ and $\nabla \times \mathbf{E}=0$, so $\nabla \times \mathbf{P}=0$ in each uniform region. The curl term vanishes and:

$$\mathbf{D}(\mathbf{r})=-\nabla \int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_f({\mathbf{r}}^{\prime })}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}=-\frac{1}{4\pi }\nabla \int {\mathrm{d}}^3{r}^{\prime }\frac{{\rho }_f({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

$\mathbf{D}$ is determined entirely by free charge — it is irrotational and thus derivable from a scalar potential. This greatly simplifies boundary value problems in simple media, but recall that surface contributions from $\nabla \times \mathbf{P}$ at interfaces can reintroduce the curl term when $ϵ$ is piecewise constant.

6.18 — Surface Polarization Charge

Point charges $q_1,q_2,\dots ,q_N$ are embedded in a body with permittivity ${\kappa }_{\text{in}}$, itself embedded in a body with permittivity ${\kappa }_{\text{out}}$. Find the total polarization charge $Q_{\text{pol}}$ on the boundary.

Solution Sketch

Let $Q_f={\sum }_k{q}_k$. The total polarization charge on the boundary $S$ is the difference of the normal electric field integrals evaluated from outside and inside:

$$Q_{\text{pol}}=-{ϵ}_0\left({\int }_{S_{\text{out}}}\mathrm{d}\mathbf{S}\cdot {\mathbf{E}}_{\text{out}}-{\int }_{S_{\text{in}}}\mathrm{d}\mathbf{S}\cdot {\mathbf{E}}_{\text{in}}\right)$$

Since $\mathbf{D}=ϵ\mathbf{E}$, both integrals can be re-expressed in terms of $\mathbf{D}$ evaluated on their respective sides of $S$. But ${\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{D}=Q_f$ regardless of which side we evaluate on (Gauss’s law for $\mathbf{D}$). Therefore:

$${ϵ}_0{\int }_{S_{\text{out}}}\mathrm{d}\mathbf{S}\cdot {\mathbf{E}}_{\text{out}}=\frac{{ϵ}_0}{{ϵ}_{\text{out}}}{Q}_f=\frac{Q_f}{{\kappa }_{\text{out}}},{ϵ}_0{\int }_{S_{\text{in}}}\mathrm{d}\mathbf{S}\cdot {\mathbf{E}}_{\text{in}}=\frac{Q_f}{{\kappa }_{\text{in}}}$$

Thus:

$$Q_{\text{pol}}=Q_f\left(\frac{1}{{\kappa }_{\text{in}}}-\frac{1}{{\kappa }_{\text{out}}}\right)$$

Note: This accounts only for the polarization charge at the boundary $S$. The bulk polarization charge in the interior (${\rho }_P=(1/{\kappa }_{\text{in}}-1){\rho }_f$) screens each embedded charge locally and is not included in $Q_{\text{pol}}$.

Source: T.P. Doerr and Y.-K. Yu, American Journal of Physics 72, 190 (2004).

Energy and Forces

6.19 — An Elastic Dielectric

A parallel-plate capacitor (area $A$, charges $\pm q$) is filled with a compressible dielectric (permittivity $ϵ$, elastic energy $U_e=\frac{1}{2}k(d-d_0{)}^2$). (a) Find the equilibrium separation $d(q)$. (b) Sketch $V(q)$ and comment on the differential capacitance $C_d(q)=\mathrm{d}q/\mathrm{d}V$.

Solution Sketch

(a) The total energy is:

$$U(q,d)=\frac{1}{2}k(d-d_0{)}^2+\frac{q^{2}d}{2ϵA}$$

Setting $\mathrm{d}U/\mathrm{d}d=0$:

$$k(d-d_0)+\frac{q^2}{2ϵA}=0⟹d(q)=d_0-\frac{q^2}{2kϵA}$$

(b) The potential difference at equilibrium:

$$V(q)=\frac{qd(q)}{ϵA}=\frac{q}{ϵA}\left(d_0-\frac{q^2}{2kϵA}\right)=\frac{q}{C_0}-\frac{q^3}{q_0^2{C}_0}$$

where $C_0=ϵA/d_0$ and $q_0^2=2kϵd_{0}A$. The differential capacitance is:

$$C_d(q)=\frac{\mathrm{d}q}{\mathrm{d}V}=\frac{C_0}{1-3q^2/q_0^2}$$

This diverges at $q=q_0/\sqrt{3}$ and becomes negative for $q>q_0/\sqrt{3}$.

Physical insight: When $C_d<0$, adding charge decreases the voltage — electrostatic attraction compresses the plates faster than charge accumulates. Beyond $q_0/\sqrt{3}$, the plates snap together (pull-in instability, the operating principle of capacitive MEMS switches).

6.21 — A Classical Meson

Model a meson as a point dipole $\mathbf{p}$ at the center of a spherical cavity (radius $R$, $\kappa =1$) in an infinite medium with $\kappa \to 0$. (a) Find $\mathbf{D}$ and $\mathbf{E}$ everywhere for finite $\kappa$. (b) Confirm that ${\mathbf{D}}_{\text{out}}=0$ and $U_E$ is finite when $\kappa =0$.

Solution Sketch

(a) Dipole $\mathbf{p}=p\hat{\mathbf{z}}$ at the center forces $\ell =1$ structure. Ansatz: ${\phi }_{\text{in}}=(p/(4\pi {ϵ}_0{r}^2)+Ar)\cos \theta$, ${\phi }_{\text{out}}=(B/r^2)\cos \theta$. Matching $\phi$ and $\kappa \partial \phi /\partial r$ at $r=R$:

$$A=-\frac{2p}{4\pi {ϵ}_0{R}^3}\cdot \frac{\kappa -1}{2\kappa +1},B=\frac{p}{4\pi {ϵ}_0}\cdot \frac{3}{2\kappa +1}$$

(b) At $\kappa =0$: ${\mathbf{D}}_{\text{out}}=\kappa {ϵ}_0{\mathbf{E}}_{\text{out}}=0$. The interior energy splits into a dipole self-energy $\sim 1/a^3$ (divergent, $R$-independent), vanishing cross terms (angular orthogonality), and a uniform-field piece $\sim p^2/({ϵ}_0{R}^3)$. Balancing the last against surface energy $U_S\propto R^2$:

$$U_{\text{tot}}\sim \frac{A}{R^3}+B{R}^2$$

This has a finite minimum — the meson is stable, in stark contrast to the free quark whose energy diverges as $\kappa \to 0$.

6.22 — An Application of the Dielectric Stress Tensor

A metal ball with charge $Q$ sits at the center of a thin spherical conducting shell (charge $Q^{\prime }$). The space between them is filled with dielectric constant $\kappa$. Show that if the shell is split into hemispheres, they stay together only if $Q$ and $Q^{\prime }$ have opposite signs and $1-1/\sqrt{\kappa }\le |Q^{\prime }/Q|\le 1+1/\sqrt{\kappa }$.

Solution Sketch

Use the stress tensor on two surfaces hugging the shell — one just inside (in the dielectric), one just outside (in vacuum).

Inner surface (radius $R^{-}$): By Gauss’s law, $D_{\text{in}}=Q/(4\pi R^2)$ and $E_{\text{in}}=D_{\text{in}}/(\kappa {ϵ}_0)$:

$$f_{\text{in}}=\frac{Q^2}{32{\pi }^2\kappa {ϵ}_0{R}^4}$$

Outer surface (radius $R^{+}$): $D_{\text{out}}=(Q+Q^{\prime })/(4\pi R^2)$:

$$f_{\text{out}}=\frac{(Q+Q^{\prime }{)}^2}{32{\pi }^2{ϵ}_0{R}^4}$$

Stability requires $f_{\text{in}}>f_{\text{out}}$, i.e., $Q^2/\kappa >(Q+Q^{\prime }{)}^2$. Setting $x=Q^{\prime }/Q$:

$$x^2-2x+1-\frac{1}{\kappa }<0$$

This parabola has zeroes at $x=1\pm 1/\sqrt{\kappa }$, giving the stability window:

$$1-\frac{1}{\sqrt{\kappa }}\le \left|\frac{Q^{\prime }}{Q}\right|\le 1+\frac{1}{\sqrt{\kappa }}$$

Since $\kappa >1$, the quadratic’s constant term is positive, forcing $Q$ and $Q^{\prime }$ to have opposite sign.

6.23 — Two Dielectric Interfaces

Two fixed-potential capacitors are filled with equal amounts of two simple dielectrics (${\kappa }_1$, ${\kappa }_2$). In one capacitor the interface is horizontal (perpendicular to $\mathbf{E}$); in the other it is vertical (parallel to $\mathbf{E}$). Use the stress tensor to compare the force per unit area on each interface.

Solution Sketch

Let $\hat{\mathbf{z}}$ point upward (along $\mathbf{E}$ in each capacitor), and choose a surface $S$ snugly enclosing each interface. The force on the interface is $\mathbf{F}={\int }_S\mathrm{d}S[(\hat{\mathbf{n}}\cdot \mathbf{D})\mathbf{E}-\frac{1}{2}\hat{\mathbf{n}}(\mathbf{E}\cdot \mathbf{D})]$.

Horizontal interface ($\hat{\mathbf{n}}=\hat{\mathbf{z}}$, $\mathbf{E}\perp$ interface): The force per unit area is:

$$f=\frac{1}{2}(E_2{D}_2-E_1{D}_1)=\frac{1}{2}{ϵ}_0({\kappa }_2{E}_2^2-{\kappa }_1{E}_1^2)$$

Since $D_z$ is continuous at the interface: $D={ϵ}_0{\kappa }_1{E}_1={ϵ}_0{\kappa }_2{E}_2$. With each dielectric layer of thickness $d/2$ and $E_0=V/d$:

$$D=\frac{2{ϵ}_0{E}_0}{1/{\kappa }_1+1/{\kappa }_2}$$ $$f=\frac{D^2}{2{ϵ}_0}\left(\frac{1}{{\kappa }_2}-\frac{1}{{\kappa }_1}\right)$$

Vertical interface ($\hat{\mathbf{n}}=\hat{\mathbf{x}}$, $\mathbf{E}\parallel$ interface): Now $\mathbf{E}=E\hat{\mathbf{z}}$ and $\hat{\mathbf{n}}=\hat{\mathbf{x}}$, so $\hat{\mathbf{n}}\cdot \mathbf{D}=D_x=0$ and $\hat{\mathbf{n}}\cdot \mathbf{E}=0$. The stress tensor gives:

$$f=-\frac{1}{2}(E_2{D}_2-E_1{D}_1)=\frac{1}{2}{ϵ}_0{E}_0^2({\kappa }_2-{\kappa }_1)$$

Here $E_1=E_2=E_0=V/d$ (tangential $\mathbf{E}$ is continuous).

Physical insight: In the vertical case, no field lines cross the interface ($\hat{\mathbf{n}}\cdot \mathbf{D}=0$), yet the $-\frac{1}{2}\hat{\mathbf{n}}(\mathbf{E}\cdot \mathbf{D})$ pressure term still produces a net force toward the lower-$\kappa$ medium.

6.25 — Minimizing the Total Energy Functional

Use Lagrange multipliers to show that, among all $\mathbf{D}(\mathbf{r})$ satisfying $\nabla \cdot \mathbf{D}={\rho }_f$, the minimum of $U_E=\frac{1}{2}\int {\mathrm{d}}^{3}r|\mathbf{D}{|}^2/ϵ$ occurs when $\mathbf{D}(\mathbf{r})=ϵ\mathbf{E}(\mathbf{r})$.

Solution Sketch

The constraint $\nabla \cdot \mathbf{D}={\rho }_f$ holds at every point, so the Lagrange multiplier is a function $\phi (\mathbf{r})$. Construct:

$$F[\mathbf{D}]=\frac{1}{2}{\int }_V{\mathrm{d}}^{3}r\frac{|\mathbf{D}{|}^2}{ϵ}-{\int }_V{\mathrm{d}}^{3}r\phi (\mathbf{r})(\nabla \cdot \mathbf{D}-{\rho }_f)$$

Integrating by parts and varying $\mathbf{D}\to \mathbf{D}+\delta \mathbf{D}$:

$$\delta F={\int }_V{\mathrm{d}}^{3}r\left(\frac{\mathbf{D}}{ϵ}+\nabla \phi \right)\cdot \delta \mathbf{D}$$

Setting $\delta F=0$ for arbitrary $\delta \mathbf{D}$:

$$\mathbf{D}=-ϵ\nabla \phi =ϵ\mathbf{E}$$

The second-order variation $\frac{1}{2}\int {\mathrm{d}}^{3}r|\delta \mathbf{D}{|}^2/ϵ>0$ confirms a true minimum. The Lagrange multiplier is the electrostatic potential itself.