Gauss's Law in Dielectrics

The previous article ended with a magic trick: define $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ , and Gauss's law reorganizes itself so that only free charge — the charge you actually deposited with a battery or a wire — appears as a source. One symmetric example (a charged wire in insulation) showed the trick works. This article is where the trick becomes a trade: the full working catalog of $\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=Q_{\text{free,enc}}$ , run on the three symmetric geometries, with every step of the bookkeeping exposed. The stakes are practical. Inside matter, the charge that sources $\mathbf{E}$ includes bound charge — and bound charge depends on the polarization, which depends on the field, which depends on the bound charge. Solving for $\mathbf{E}$ head-on means untangling that loop. The displacement field sidesteps it: in a symmetric problem, $\mathbf{D}$ follows from the free charge in one line, before you know anything about the material's response. The polarization, the field, and the bound charges are then harvested afterwards, in order, like reading answers off the back of the book. We run the ritual on a flagship problem — a point charge wrapped in a polarized spherical shell — twice: once the honest, laborious way (find every bound charge, then use ordinary Gauss's law), once the $\mathbf{D}$ way. Watching half a page collapse into two lines is the whole argument for $\mathbf{D}$ 's existence. And because the shortcut has a failure mode, we also stage its collapse: a geometry where the free charge is zero, the flux of $\mathbf{D}$ is zero, and $\mathbf{D}$ itself is stubbornly not zero.

Key takeaways

$\nabla\cdot\mathbf{D}=\rho_{\text{free}}$ and its integral twin $\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=Q_{\text{free,enc}}$ (established in The Electric Displacement D) turn symmetric dielectric problems into one-line flux algebra.
The ritual: check symmetry → get $\mathbf{D}$ from free charge alone → recover $\mathbf{E}=(\mathbf{D}-\mathbf{P})/\varepsilon_0$ → harvest $\sigma_b$ , $\rho_b$ → audit that the bound charges total zero.
$\mathbf{D}$ is blind to the dielectric: for a point charge in a polarized shell, $\mathbf{D}=q\hat{\mathbf{r}}/4\pi r^2$ everywhere, shell or no shell. All the material's drama lives in $\mathbf{E}$ .
The same answer emerges from the long route — sum free plus bound charge and apply ordinary Gauss's law for $\mathbf{E}$ — and the two routes cross-check each other exactly.
The shortcut needs symmetry: with $\rho_{\text{free}}=0$ , the flux of $\mathbf{D}$ vanishes through every closed surface, yet $\mathbf{D}\neq0$ (a frozen-polarization electret proves it). Flux constrains totals, not fields.
KEY IDEA: $\nabla\cdot\mathbf{D}=\rho_{\text{free}}$ packages all bound-charge effects inside $\mathbf{D}$ , so in symmetric geometries the displacement follows from free charge alone via $\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=Q_{\text{free,enc}}$ — but symmetry, not the law by itself, is what makes $\mathbf{D}$ computable.

Where this lives today

Every semiconductor device simulator on Earth spends its inner loop solving equation (1). TCAD tools discretize $\nabla\cdot(\varepsilon\nabla V)=-\rho_{\text{free}}$ over transistor geometries with a dozen different dielectrics — gate oxides, high- $\kappa$ hafnia stacks, interlayer insulators — precisely because free charge (dopants, channel carriers) is what the engineer specifies, while the bound response rides inside $\varepsilon$ . The same free-charge-only bookkeeping runs the design of high-voltage DC cables, with a modern twist this article's Move 4 anticipates: in aged polymer insulation, slow charge injection creates genuine free space charge inside the jacket, and the measured field distortion is diagnosed by exactly the coax audit of worked example 2 — flux the measured charge, compare against the clean $\lambda_f/2\pi s$ baseline. The electret derivation is not a curiosity either: electret microphones (a frozen-polarization film generating a permanent bias field, no battery needed) sit in billions of phones and laptops, and their design field is computed by the boundary-value route precisely because — as Derivation C staged — no Gaussian surface will hand it to you. Knowing when the shortcut applies, and when it silently doesn't, is the practical skill this catalog trains.

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen Observatory

Gauss never wrote $\mathbf{D}$ — the displacement is Maxwell's coinage — but both halves of this article's machinery are his. The divergence theorem that converts equation (1) into the flux law (2) appears in his 1813 memoir on the attraction of spheroids, and the flux–charge relation itself he stated in an 1835 manuscript (published only posthumously, in 1867 — Gauss hoarded results the way other men hoard money, his motto pauca sed matura, "few but ripe"). Fittingly for this article, his deepest engagement with real measurements-in-media came through fields in matter of another kind: with Wilhelm Weber he built the Göttingen magnetic observatory, organized the worldwide Magnetischer Verein, and invented the systematic separation of a measured field into its sources — the ancestor of every free-versus-bound decomposition, including the one $\mathbf{D}$ performs. The flux ritual this article drills — pick the symmetric surface, count only the sources you control — is the Gaussian habit of mind: reduce a field problem to an accounting identity, then let geometry do the labor.

History

1813

— Gauss's spheroid memoir develops the divergence theorem, the bridge between the differential and integral laws

1835

— Gauss states the flux–charge law in manuscript; published posthumously in 1867

1837

— Faraday's dielectric experiments show matter reshapes capacitor fields, creating the need for two-field bookkeeping

1864

— Maxwell introduces the displacement and writes Gauss's law in matter in its modern source-split form

1885

— Heaviside recasts Maxwell's equations in vector form; div D equals rho-free becomes one of the canonical four

The physical picture

Think of an electrostatics problem in matter as a business with two ledgers. The free-charge ledger records what the experimenter paid in: electrons pushed onto a plate, ions implanted in a crystal, a line charge run down a cable. The bound-charge ledger records the medium's internal shuffling — dipole heads and tails that fail to cancel at surfaces and wherever the polarization is uneven (the accounting rules $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}$ and $\rho_b=-\nabla\cdot\mathbf{P}$ from Bound Charges: Surface and Volume). The field $\mathbf{E}$ answers to both ledgers. The displacement $\mathbf{D}$ answers only to the first — and the first ledger is the one you can read without solving anything, because you wrote it yourself.

Figure

One free charge, two fields — D ignores the shell, E feels it

The picture above is the article in miniature. A free point charge sits in a cavity inside a polarized shell. The flux function $4\pi r^2 D$ is flat — a horizontal line at the value $q$ , sailing through cavity, shell, and vacuum without noticing them, because no surface encloses any free charge but $q$ . The corresponding plot for $\varepsilon_0\mathbf{E}$ dips and jumps: down at the inner surface (negative bound charge), sagging through the shell (volume bound charge), back up at the outer surface (positive bound charge), and rejoining the $\mathbf{D}$ line outside — where the bound ledger, summed, reads zero, as it always must for a neutral dielectric. One caution before the catalog, inherited from The Electric Displacement D and enforced throughout: $\mathbf{D}$ obeys a Gauss law but not a Coulomb law. Its curl, $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$ , is generally nonzero, so free charge alone does not determine $\mathbf{D}$ — the flux law pins down $\mathbf{D}$ only when symmetry has already dictated its direction and its dependence on one coordinate. Every entry in the catalog below begins by checking that license. The final derivation stages what happens when the license is missing.

The law and the ritual

The tools, both established in The Electric Displacement D and recalled here rather than re-derived. In differential form, with $\mathbf{D}\equiv\varepsilon_0\mathbf{E}+\mathbf{P}$ absorbing the bound charge $\rho_b=-\nabla\cdot\mathbf{P}$ into the field's definition, Gauss's law in matter reads

\nabla\cdot\mathbf{D}=\rho_{\text{free}} \qquad (1)

where: $\mathbf{D}$ — electric displacement (C/m²); $\rho_{\text{free}}$ — density of the charge you control (C/m³). The bound charge has not disappeared; it is hiding inside $\mathbf{D}$ 's definition, which is exactly where you want it while solving. Integrating (1) over a volume and applying the divergence theorem gives the workhorse form. If you are choosing a Gaussian surface, this is the equation you are about to use:

\oint_{S}\mathbf{D}\cdot\mathrm{d}\mathbf{a}=Q_{\text{free,enc}} \qquad (2)

where: $S$ — any closed surface; $Q_{\text{free,enc}}$ — the free charge inside it, only the free charge. Compare the vacuum law from Gauss's Law in Integral Form: same geometry, same symmetry arguments, but the right-hand side no longer asks you to know the medium's response. That ignorance-tolerance is the entire selling point. Once $\mathbf{D}$ is known, everything else unwinds by definition — no material model needed if $\mathbf{P}$ is given:

\mathbf{E}=\frac{\mathbf{D}-\mathbf{P}}{\varepsilon_0} \qquad (3)

where: $\mathbf{P}$ — the polarization, prescribed in this article's problems. (When instead the material responds linearly, $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$ and (3) becomes $\mathbf{E}=\mathbf{D}/\varepsilon$ — the constitutive machinery of the next article, Linear Dielectrics, Susceptibility and Permittivity, previewed here and deferred there.) The catalog procedure — the ritual — is the same five moves every time.

Move 1. Certify symmetry: planar, cylindrical, or spherical free charge and material arrangement, so that $\mathbf{D}$ must point along the symmetry direction and depend on one coordinate only.

Move 2. Run equation (2) on the symmetry-adapted surface; solve algebraically for $D$ .

Move 3. Recover $\mathbf{E}$ from (3), region by region — this is where the material finally enters.

Move 4. Harvest the bound charges: $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}$ on every surface, $\rho_b=-\nabla\cdot\mathbf{P}$ in every volume (rules owned by Bound Charges: Surface and Volume).

Move 5. Audit: the bound charges of a neutral dielectric must total zero, and ordinary Gauss's law for $\mathbf{E}$ , run with free plus bound charge, must reproduce Move 3's field. If it doesn't, a sign error is hiding in Move 4.

geometry	Gaussian surface	Move 2 output
planar (sheets, slabs, plates)	pillbox straddling the structure	$D=\sigma_f$ (from the charged plate)
cylindrical (wires, coax, jackets)	coaxial cylinder, radius $s$ , length $L$	$D=\lambda_f/2\pi s$
spherical (point charges, shells)	concentric sphere, radius $r$	$D=Q_f/4\pi r^2$

The table is deliberately short. Three rows is the entire list of geometries where equation (2) determines $\mathbf{D}$ outright — the same three as in vacuum (Gauss's Law: Applications and Symmetry), for the same reason: only there does symmetry pin the field's direction before any equation is solved.

Figure

The five-move ritual — free charge in, audited fields out

The flagship, twice

The problem: a free point charge $q$ at the center of a thick dielectric shell (inner radius $a$ , outer radius $b$ ) with frozen-in polarization $\mathbf{P}=(k/r)\,\hat{\mathbf{r}}$ inside the shell, vacuum elsewhere. Find $\mathbf{E}$ everywhere. We solve it both ways, deliberately in this order: the honest slog first, so the shortcut has something to be compared against.

Derivation ARoute one, the honest slog — every bound charge, then ordinary Gauss

We refuse the shortcut: catalog every scrap of bound charge the polarization creates, then apply the ordinary vacuum Gauss's law with the total (free plus bound) enclosed charge. The strategy is pure bookkeeping — the accounting rules for bound charge, applied to three locations (inner surface, shell volume, outer surface) — and the main thing to watch is signs: the inner surface's outward normal points into the cavity, toward $-\hat{\mathbf{r}}$ . The payoff is a complete audit trail that the fast route must then match.

Step 1. Volume bound charge. In spherical coordinates the divergence of a radial field $P(r)\hat{\mathbf{r}}$ is $(1/r^2)\,\partial_r(r^2P)$ , so with $P=k/r$ :

\rho_b=-\nabla\cdot\mathbf{P}=-\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\cdot\frac{k}{r}\right)=-\frac{1}{r^2}\frac{\partial}{\partial r}(kr)=-\frac{k}{r^2}.

Why negative: $\mathbf{P}$ weakens as $1/r$ going outward, but the area of each spherical shell grows as $r^2$ , so $r^2P\propto r$ still grows — dipole heads keep outrunning tails, leaving net negative charge behind in every shell of the volume.

Step 2. Surface bound charges. At the outer surface $\hat{\mathbf{n}}=+\hat{\mathbf{r}}$ , so $\sigma_b(b)=\mathbf{P}\cdot\hat{\mathbf{n}}=+k/b$ . At the inner surface the outward normal of the dielectric points toward the cavity, $\hat{\mathbf{n}}=-\hat{\mathbf{r}}$ , so $\sigma_b(a)=-k/a$ .

Step 3. Audit the totals before using them. Inner surface: $(-k/a)(4\pi a^2)=-4\pi ka$ . Volume: $\int_a^b(-k/r^2)\,4\pi r^2\,\mathrm{d}r=-4\pi k(b-a)$ . Outer surface: $(+k/b)(4\pi b^2)=+4\pi kb$ . Sum: $-4\pi ka-4\pi kb+4\pi ka+4\pi kb=0$ . Why this check matters: a neutral dielectric shuffles charge but cannot create it — the theorem proved in Bound Charges: Surface and Volume. Any nonzero total here means a sign slipped in Step 1 or 2.

Step 4. Enclosed charge as a function of $r$ , then ordinary Gauss's law $\oint\varepsilon_0\mathbf{E}\cdot\mathrm{d}\mathbf{a}=Q_{\text{enc}}$ . In the cavity ( $r<a$ ): $Q_{\text{enc}}=q$ , so $E=q/4\pi\varepsilon_0r^2$ — the bare Coulomb field. Inside the shell ( $a<r<b$ ): add the inner surface and the volume down to $r$ ,

Q_{\text{enc}}(r)=q-4\pi ka-4\pi k(r-a)=q-4\pi kr \quad\Longrightarrow\quad E(r)=\frac{q}{4\pi\varepsilon_0 r^2}-\frac{k}{\varepsilon_0 r}.

Outside ( $r>b$ ): the bound ledger closes to zero (Step 3), $Q_{\text{enc}}=q$ , and the bare Coulomb field returns as if the shell were never there. Why the exterior field recovers the bare Coulomb form: this is forced by Step 3's zero total — the bound ledger closes to nothing outside the shell, so the shell dents the field only where you stand inside it.

Go deeper · Derivation B · 1 minRoute two, the displacement shortcut — and why it is two lines

Derivation BRoute two, the displacement shortcut — and why it is two lines

Same problem, now with equation 2. The strategy: spherical symmetry certifies that $\mathbf{D}=D(r)\hat{\mathbf{r}}$ ; the flux law then reads off $D$ from the only free charge in the problem; equation 3 converts to $\mathbf{E}$ region by region. Watch how the polarization — the entire subject of route one — appears exactly once, in the last line, as an afterthought.

Step 1. Everything in sight (charge, shell, polarization pattern) is spherically symmetric, so $\mathbf{D}$ is radial and depends only on $r$ . A concentric Gaussian sphere of any radius $r$ gives

\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=4\pi r^2D=Q_{\text{free,enc}}=q \quad\Longrightarrow\quad \mathbf{D}=\frac{q}{4\pi r^2}\,\hat{\mathbf{r}}\;\;\text{everywhere}.

Why "everywhere" is the remarkable word: cavity, shell, and vacuum all get the same formula, because no Gaussian sphere ever encloses free charge other than $q$ . The bound charges that route one tabulated so carefully are invisible to this equation by construction.

Step 2. Unwind $\mathbf{E}$ from equation 3. In the cavity and outside, $\mathbf{P}=0$ , so $\mathbf{E}=\mathbf{D}/\varepsilon_0=q\hat{\mathbf{r}}/4\pi\varepsilon_0r^2$ . Inside the shell,

\mathbf{E}=\frac{\mathbf{D}-\mathbf{P}}{\varepsilon_0}=\left(\frac{q}{4\pi\varepsilon_0 r^2}-\frac{k}{\varepsilon_0 r}\right)\hat{\mathbf{r}},

identical to route one's Step 4 — every region, every term. Why the two routes agree exactly — and why this is not a coincidence but an identity: equation 1 is simply Gauss's law for $\mathbf{E}$ with the bound charge moved across the equals sign. Route two is route one with the bookkeeping pre-paid.

The comparison deserves one paragraph of savoring. Route one needed a divergence, two normal-vector sign decisions, a three-part audit, and an integral. Route two needed the sentence "spherical symmetry, free charge $q$ ." Both are correct; only one scales. When the geometry is symmetric, always take the $\mathbf{D}$ route and use route one's machinery as the audit (Moves 4–5 of the ritual). This division of labor — $\mathbf{D}$ for solving, bound charges for understanding and checking — is how every dielectric problem in the rest of the track is handled.

Go deeper · Derivation C · 2 minWhen the license is missing — zero flux with a nonzero field

Derivation CWhen the license is missing — zero flux with a nonzero field

Now the failure mode, staged deliberately. Take a bar electret: a cylinder of material with uniform frozen polarization $\mathbf{P}=P_0\hat{\mathbf{z}}$ , no free charge anywhere (the "deceptive parallel" warning of The Electric Displacement D, now with the flux argument run in full). We show that every closed surface reports zero flux of $\mathbf{D}$ , and yet $\mathbf{D}$ vanishes nowhere near the bar — so the flux law, with nothing else, determines nothing.

Step 1. The flux audit. With $\rho_{\text{free}}=0$ identically, equation 2 gives $\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=0$ through every closed surface — around the whole bar, through its middle, anywhere.

Step 2. The field audit. The electret's bound charges ( $\sigma_b=\pm P_0$ on the end faces) produce the field of a capacitor-like charge pair: $\mathbf{E}$ points tail-to-head outside the bar and head-to-tail inside, exactly like a bar magnet's $\mathbf{H}$ -pattern — it is $\mathbf{D}$ , not $\mathbf{E}$ , that plays $\mathbf{B}$ 's role here. Then $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ : outside the bar $\mathbf{D}=\varepsilon_0\mathbf{E}\neq0$ ; inside, near the middle, $\varepsilon_0|\mathbf{E}|<P_0$ (the depolarizing field of two distant charged disks is weaker than $P_0/\varepsilon_0$ ), so $\mathbf{D}=\mathbf{P}+\varepsilon_0\mathbf{E}\neq0$ there too — $\mathbf{D}$ 's field lines form closed loops through the bar. Why loops are the tell: closed field lines mean zero net flux through any closed surface (consistent with Step 1) but a nonzero curl. Indeed $\nabla\times\mathbf{D}=\varepsilon_0\underbrace{\nabla\times\mathbf{E}}_{0}+\nabla\times\mathbf{P}$ , and at the bar's side walls $\mathbf{P}$ snaps from $P_0\hat{\mathbf{z}}$ to zero — a discontinuity with a curl. $\mathbf{D}$ is sourced by free charge and curled by polarization edges.

Step 3. The moral, stated with care. A vector field is determined by its divergence and its curl (with boundary data) — the Helmholtz theorem from the E0 chapter. Equation 1 supplies only the divergence; symmetric problems survive because symmetry substitutes for the missing curl information. The electret has no such symmetry axis of the right kind: no Gaussian surface exists on which $\mathbf{D}$ is constant-magnitude and normal, so Move 1 of the ritual fails and the flux law, though true, is useless for finding $\mathbf{D}$ . One must instead solve the boundary-value problem for the potential — the machinery of Boundary Conditions for D and E and, for spheres, the Dielectric Sphere in a Uniform Field. Why the curl identity is immediate: $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$ follows in one line from the definition $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ together with the electrostatic irrotationality $\nabla\times\mathbf{E}=0$ .

Limiting cases

Case	Result	Meaning
No matter anywhere	$\mathbf{P}=0\Rightarrow\mathbf{D}=\varepsilon_0\mathbf{E}$	equation (2) reduces to the vacuum Gauss's law — same three geometries, same ritual
Shell with $q$ at center, $r>b$	$E=q/4\pi\varepsilon_0r^2$ exactly	bound charge totals zero; from outside, a neutral dielectric is electrically invisible
Shell region, $\mathbf{P}=(k/r)\hat{\mathbf{r}}$	$E=q/4\pi\varepsilon_0r^2-k/\varepsilon_0r$	the medium's dent in $\mathbf{E}$ ; $\mathbf{D}$ unchanged
Uniform $\mathbf{P}$ , no free charge (electret)	$\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=0$ but $\mathbf{D}\neq0$	flux law true yet useless; symmetry license missing
Linear medium (next article)	$\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}\Rightarrow\mathbf{E}=\mathbf{D}/\varepsilon$	Move 3 becomes a division; the catalog re-runs with $\varepsilon$ doing the work

Common trapZero enclosed free charge does not mean D is zero

The single most common misreading of equation (2). "No free charge inside my surface, so $\oint\mathbf{D}\cdot\mathrm{d}\mathbf{a}=0$ , so $\mathbf{D}=0$ " — the first two clauses are right and the third is a non sequitur, exactly as "zero flux of $\mathbf{E}$ " never meant "zero $\mathbf{E}$ " back in Electric Flux. Zero flux constrains the total, not the field pointwise; the electret derivation exhibits a $\mathbf{D}$ that threads closed loops through every zero-flux surface. The flux law hands you $\mathbf{D}$ only after symmetry has certified that $\mathbf{D}$ is normal to your Gaussian surface with uniform magnitude — Move 1 is not a formality, it is the license for everything that follows. No symmetry, no shortcut: solve the boundary-value problem instead.

Worked examples

1 · A polarized slab parked in a charged capacitor (confidence builder)

Parallel plates carry free charge $\pm\sigma_f=\pm8.854\,\mu\mathrm{C/m^2}$ , separated by $d=2.0\,\mathrm{mm}$ . A slab with frozen uniform polarization $P=4.427\,\mu\mathrm{C/m^2}$ (pointing along $\mathbf{E}$ , i.e. from the positive plate toward the negative) and thickness $t=1.0\,\mathrm{mm}$ sits against one plate; the remaining $1.0\,\mathrm{mm}$ is vacuum. Find $\mathbf{D}$ , $\mathbf{E}$ in both regions, the bound charges, and the plate voltage.

Step 1. Ritual Move 1–2. Planar symmetry throughout (plates and slab are uniform in the transverse directions). A pillbox with one face buried in the positive plate's conductor and one face in the gap encloses free charge $\sigma_f$ per unit area, so $D=\sigma_f=8.854\,\mu\mathrm{C/m^2}$ — and since no free charge lives anywhere between the plates, this same $D$ holds in the vacuum gap and inside the slab.

Step 2. Move 3, region by region. Vacuum gap: $E_{\text{gap}}=D/\varepsilon_0=(8.854\times10^{-6})/(8.854\times10^{-12})=1.00\,\mathrm{MV/m}$ . Slab: $E_{\text{slab}}=(D-P)/\varepsilon_0=(8.854-4.427)\times10^{-6}/(8.854\times10^{-12})=0.50\,\mathrm{MV/m}$ — half the gap field, because the polarization shoulders half the displacement.

Step 3. Move 4–5, harvest and audit. Slab faces carry $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}=\mp4.427\,\mu\mathrm{C/m^2}$ (negative on the face toward the positive plate). Audit by superposition of charged sheets: near the positive plate the slab's two bound faces together cancel half of $\sigma_f$ 's push inside the slab, giving $(\sigma_f-P)/\varepsilon_0=0.50\,\mathrm{MV/m}$ — matching Step 2 exactly. Total bound charge: two equal-and-opposite faces, zero.

Step 4. Voltage: $V=E_{\text{gap}}(d-t)+E_{\text{slab}}\,t=(1.00\times10^6)(1.0\times10^{-3})+(0.50\times10^6)(1.0\times10^{-3})=1500\,\mathrm{V}$ , versus $2000\,\mathrm{V}$ for the empty capacitor at the same free charge. The slab eats $500\,\mathrm{V}$ of the voltage at fixed charge — the germ of why dielectrics raise capacitance, quantified properly in the next article.

2 · Coax cable with a polarized jacket (exam level)

A long inner conductor of radius $a=1.0\,\mathrm{mm}$ carries free line charge $\lambda_f=50\,\mathrm{nC/m}$ . Its insulating jacket ( $a<s<b$ , $b=3.0\,\mathrm{mm}$ ) carries frozen polarization $\mathbf{P}=(\alpha/s)\,\hat{\mathbf{s}}$ with $\alpha=2.0\,\mathrm{nC/m}$ . Find the fields, all bound charge, and the voltage across the jacket — then audit by the long route.

Step 1. Moves 1–2. Cylindrical symmetry; a coaxial Gaussian cylinder of radius $s$ and length $L$ encloses free charge $\lambda_fL$ regardless of $s>a$ , so $D(2\pi sL)=\lambda_fL$ gives $D=\lambda_f/2\pi s$ . At the inner surface $D(a)=50\times10^{-9}/(2\pi\times10^{-3})=7.96\,\mu\mathrm{C/m^2}$ , where the polarization is $P(a)=\alpha/a=2.00\,\mu\mathrm{C/m^2}$ .

Step 2. Move 3. Inside the jacket, $E=(D-P)/\varepsilon_0=\left(\lambda_f/2\pi-\alpha\right)/\varepsilon_0s=672.9/s\ \mathrm{V/m}$ ( $s$ in meters): $E(a)=0.673\,\mathrm{MV/m}$ at the conductor, falling to $E(b)=224\,\mathrm{kV/m}$ at the outer edge. Outside the jacket $P=0$ and the vacuum form $E=\lambda_f/2\pi\varepsilon_0s$ resumes.

Step 3. Move 4. Volume: $\rho_b=-\nabla\cdot\mathbf{P}=-(1/s)\,\partial_s(s\cdot\alpha/s)=-(1/s)\,\partial_s\alpha=0$ — the $1/s$ profile is precisely the one with no volume bound charge (its $s P$ is constant, the cylindrical analogue of the shell's trick question). Surfaces: $\sigma_b(a)=-\alpha/a=-2.00\,\mu\mathrm{C/m^2}$ , $\sigma_b(b)=+\alpha/b=+0.667\,\mu\mathrm{C/m^2}$ .

Step 4. Move 5, the audit. Bound charge per unit length: $2\pi a\,\sigma_b(a)+2\pi b\,\sigma_b(b)=-2\pi\alpha+2\pi\alpha=0$ . Long route inside the jacket: enclosed line charge $=\lambda_f-2\pi\alpha=50-12.57=37.4\,\mathrm{nC/m}$ , so $E=\lambda_{\text{enc}}/2\pi\varepsilon_0s=672.9/s\,\mathrm{V/m}$ — identical to Step 2.

Step 5. Voltage across the jacket: $V=\int_a^b E\,\mathrm{d}s=672.9\,\ln(b/a)=672.9\times\ln 3=739\,\mathrm{V}$ , versus $987\,\mathrm{V}$ for the same free charge with a vacuum gap. The jacket's inner negative bound skin ( $-2.00\,\mu\mathrm{C/m^2}$ ) screens a quarter of the line charge and trims $248\,\mathrm{V}$ — real cable insulation does the same job with linear response instead of frozen $\mathbf{P}$ , the case the catalog re-runs next article.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §4.3 · Jackson, Classical Electrodynamics, 3rd ed., §4.3 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 10 · Zangwill, Modern Electrodynamics, §6.3 Key relations used throughout: $\nabla\cdot\mathbf{D}=\rho_{\text{free}}$ follows directly from $\varepsilon_0\nabla\cdot\mathbf{E}=\rho_{\text{free}}-\nabla\cdot\mathbf{P}$ ; the filled-capacitor results are $C=\varepsilon_r\varepsilon_0 A/d$ and $\sigma_b=-\sigma_f(1-1/\varepsilon_r)$ ; the screened point-charge field is $E=q/4\pi\varepsilon_0\varepsilon_r r^2$ (spherical Gaussian surface); and $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$ . Throughout, $\varepsilon_0=8.854\times10^{-12}$ F/m (CODATA).

Builds on: Polarization Density P · Bound Charges: Surface and Volume · The Electric Displacement D · Dielectric Materials and Insulators

Leads to: Linear Dielectrics, Susceptibility and Permittivity · Boundary Conditions for D and E · Dielectric Sphere in a Uniform Field

If D is the same with or without the shell, why bother with the dielectric at all?

Because you live in $\mathbf{E}$ , not $\mathbf{D}$ . Forces on charges, voltages between conductors, breakdown thresholds — all are set by $\mathbf{E}$ , and the shell very much changes $\mathbf{E}$ (it trimmed $739$ versus $987\,\mathrm{V}$ in the coax example). $\mathbf{D}$ 's invariance is not the physics; it is the solving strategy. $\mathbf{D}$ is the quantity that stays computable while the material does complicated things, which is exactly what makes it the right first thing to calculate and the wrong thing to stop at.

Why does only free charge count in the flux of D when bound charge is just as real?

Bound charge is completely real — it deflects fields, does work, shocks fingers. The point is bookkeeping, not ontology: $\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$ is defined so that the $\mathbf{P}$ term's divergence exactly cancels $\rho_b$ in Gauss's law. The bound charge hasn't been dismissed; it has been folded into the field variable itself. When you compute $\mathbf{E}=(\mathbf{D}-\mathbf{P})/\varepsilon_0$ in Move 3, you are unfolding it again — and Move 5's audit (rerun Gauss for E with all the charge) checks that the fold was clean.

How do I know which surfaces get sigma-b and what sign to use?

Every surface where $\mathbf{P}$ ends or changes discontinuously gets a $\sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}}$ , with $\hat{\mathbf{n}}$ the outward normal of the dielectric — the convention from Bound Charges: Surface and Volume. The classic trap is an inner cavity surface, where "outward from the dielectric" points toward the cavity center: that is why the shell picked up $\sigma_b(a)=-k/a$ , not $+k/a$ . If your signs are right, the total bound charge of a neutral dielectric comes out zero; if it doesn't, hunt for a flipped normal before doubting anything else.

Is D just epsilon-zero E in the cavity and outside the shell? Why?

Yes — wherever $\mathbf{P}=0$ (vacuum, air to excellent approximation, the cavity, the region beyond $b$ ), the definition collapses to $\mathbf{D}=\varepsilon_0\mathbf{E}$ and the two fields are the same object in different units. That is also why the $4\pi r^2\varepsilon_0E$ curve in the hero figure rejoins the flat $\mathbf{D}$ line outside the shell: with zero total bound charge and no polarization out there, nothing distinguishes the two bookkeepings anymore.

The electret has no symmetry for the flux law, but the polarized slab in example 1 also had no free charge inside the gap. Why did the slab work?

Because the whole problem — plates plus slab — retained planar symmetry, and the plates supplied free charge for the pillbox to see. Symmetry is a property of the entire charge-and-material arrangement, not of the dielectric alone. The slab rode on the plates' symmetry: $D=\sigma_f$ was fixed by the plate charge, and the slab merely reapportioned it between $\varepsilon_0\mathbf{E}$ and $\mathbf{P}$ . Strip the plates away and you have a free-standing electret slab — and then, like the bar, the flux law goes silent and you must solve the potential problem.

When P depends on E, as in real insulation, doesn't the ritual break because I can't do Move 3 without knowing E?

The ritual survives; only Move 3 changes character. For linear media $\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$ , so $\mathbf{D}=\varepsilon_0(1+\chi_e)\mathbf{E}\equiv\varepsilon\mathbf{E}$ , and Move 3 becomes a division by $\varepsilon$ instead of a subtraction of a known $\mathbf{P}$ . The beauty is that Moves 1–2 are untouched: $\mathbf{D}$ never cared about the material. That is the program of the next article, Linear Dielectrics, Susceptibility and Permittivity — same catalog, self-adjusting medium.

Why is there no potential for D the way V works for E?

A scalar potential exists exactly for curl-free fields, and $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$ is generally nonzero — the electret's side walls are a concrete counterexample, and Derivation C's closed $\mathbf{D}$ -loops are what a curl looks like in a picture. This is the structural asymmetry between the two fields: $\mathbf{E}$ has a potential but sees all charge; $\mathbf{D}$ sees only free charge but has no potential. Each field owns exactly one convenience, never both — The Electric Displacement D calls this the deceptive parallel, and it is the honest summary of the whole chapter.