Uniformly Polarized Sphere

Every chapter has its set-piece, and for electric fields in matter it is this one: the uniformly polarized sphere, Griffiths's Example 4.2 and Example 4.3 rolled into a single construction, the problem every later result in the chapter quietly leans on. Take a sphere of radius $R$ , freeze in a uniform polarization $\mathbf{P} = P\hat{\mathbf{z}}$ , and ask for the field everywhere. The bound-charge machinery answers instantly at the level of sources: no volume charge ( $\mathbf{P}$ uniform), and a surface charge $\sigma_b = P\cos\theta$ — positive cap, negative cap, fading to zero at the equator. The field of that cosine shell is the question, and its answer is a small miracle twice over. Inside: the field is perfectly uniform, $\mathbf{E}_{\rm in} = -\mathbf{P}/3\varepsilon_0$ — not approximately, not near the center, but exactly, everywhere within the sphere, pointing against the polarization. Of all shapes, only the ellipsoids (sphere chief among them) polarize this cleanly, and the sphere's $\tfrac13$ is the depolarizing factor that the local-field and Clausius–Mossotti constructions of this chapter will spend later articles cashing. Outside: the field is a pure dipole field, exactly, at every exterior point down to the surface — the sphere with moment $\mathbf{p} = \tfrac43\pi R^3\mathbf{P}$ impersonates a point dipole with zero error, cousin to the shell theorem's point-charge impersonation. We earn the result by two independent routes, each owning a different insight. The displaced-spheres construction — two overlapping balls of charge, shifted a hair — gets both miracles in three lines from the interior-linearity of the uniform ball, and explains why they happen. The boundary-value route treats $\sigma_b = P\cos\theta$ with the Legendre machinery of the separation-of-variables article, and shows the miracle from the other side: $\cos\theta$ is $P_1$ , a single rung of the Legendre ladder, so the response is single-rung too. One problem, two methods, three articles' worth of payoff.

Key takeaways

Uniform $\mathbf{P} = P\hat{\mathbf{z}}$ in a sphere ⇒ $\rho_b = 0$ and $\sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}} = P\cos\theta$ : charge caps of opposite sign, zero at the equator.
Inside, the field is exactly uniform: $\mathbf{E}_{\rm in} = -\mathbf{P}/3\varepsilon_0$ — the sphere's depolarizing factor is exactly $\tfrac13$ , between the needle's 0 and the slab's 1.
Outside, the field is exactly that of a point dipole $\mathbf{p} = \tfrac43\pi R^3\,\mathbf{P}$ at the center — polarization density times volume, with no higher multipoles at any distance.
Route one (displaced spheres): two uniform balls $\pm\rho$ offset by $\mathbf{d}$ overlap into a shell of $\cos\theta$ charge; the interior field is the difference of two linear ball fields — automatically uniform.
Route two (Legendre matching): $\cos\theta = P_1(\cos\theta)$ drives only the $l = 1$ term of the spherical-harmonic solution — one-rung source, one-rung response.
Boundary audit: tangential $\mathbf{E}$ continuous everywhere; normal jump $= \sigma_b/\varepsilon_0 = P\cos\theta/\varepsilon_0$ — the E1 boundary conditions verified on a live example.
KEY IDEA: the uniformly polarized sphere's $\cos\theta$ surface charge produces an exactly uniform interior field $-\mathbf{P}/3\varepsilon_0$ and an exactly dipolar exterior field with $\mathbf{p} = \tfrac43\pi R^3\mathbf{P}$ — the chapter's canonical construction and the source of the ubiquitous factor $\tfrac13$ .

The core insight

A uniformly polarized sphere carries surface charge $\sigma_b = P\cos\theta$ and no volume charge; that single-harmonic source makes the interior field exactly uniform, $\mathbf{E}_{\rm in} = -\mathbf{P}/3\varepsilon_0$ , and the exterior field exactly that of a point dipole $\mathbf{p} = \tfrac43\pi R^3\mathbf{P}$ .

Where this lives today

The sphere's $\tfrac13$ is one of the hardest-working numbers in materials physics. The Lorentz local-field construction — carve a mesoscopic spherical cavity around a molecule and ask what field it really feels — turns on the cavity-shaped twin of this article's result, and through it the Clausius–Mossotti relation connects molecular polarizability to measurable permittivity (both articles ahead in this chapter). Magnetism runs the identical mathematics: a uniformly magnetized sphere has uniform interior $\mathbf{B}$ -field structure with the same $\tfrac13$ demagnetizing factor, which is why MRI physicists shim magnets against the fields of the patient's own polarized tissue and why paleomagnetists model magnetized grains as Poisson's equivalent spheres. The exterior half — sphere as exact point dipole — is the workhorse of small-particle electromagnetics. A dielectric or metallic nanoparticle in a light wave polarizes almost uniformly (the quasi-static regime: particle much smaller than the wavelength), so it radiates and scatters as a single dipole whose strength carries the $\tfrac13$ geometry — the Rayleigh scattering that blues the sky, the localized plasmon resonance that reddens gold colloids (the metal's negative permittivity fighting the depolarizing $\tfrac13$ produces a resonance), and the biosensor color shifts that report molecular binding on nanoparticle surfaces. Effective-medium theories (Maxwell Garnett and descendants) average many such polarized spheres into designer permittivities for metamaterials — every one of them the uniformly polarized sphere, deployed at industrial scale.

Origins · 1 minLord Kelvin and how this was discovered

Lord Kelvin

1824–1907 · Glasgow

William Thomson — Kelvin from 1892 — did more than anyone to make polarized and magnetized matter calculable. His 1840s–50s papers on the "mathematical theory of magnetism" took Poisson's equivalent distributions and built the working theory of bodies with frozen or induced moments: he introduced the very terms permeability and susceptibility, proved the uniformly magnetized sphere equivalent to a point dipole (this article's exterior miracle, in magnetic dress), and — most subtly — showed that "the field inside matter" needs an operational definition, proposing the famous cavity constructions: carve a needle-shaped cavity and measure one field, a disk-shaped cavity and measure another, the difference being exactly the depolarizing physics of this article's shape hierarchy (needle 0, slab 1, sphere $\tfrac13$ ). It was Kelvin, too, who rescued Green's Essay from oblivion in 1845 and transmitted potential theory to Maxwell, whose Treatise then systematized the polarized sphere for both electricity and magnetism. Prodigy (Glasgow professor at 22), engineer (the Atlantic cable made him wealthy and knighted), absolutist about temperature (the kelvin is his), he wired together the mathematics this whole chapter runs on.

History

1600

— Gilbert's terrella, a lodestone sphere modeling the Earth, is the uniformly magnetized sphere as experiment

1824

— Poisson's equivalent distributions make magnetized (and later polarized) spheres a potential-theory problem

1851

— Kelvin's magnetism papers define susceptibility and permeability and settle fields-inside-matter via cavity constructions

1873

— Maxwell's Treatise presents the polarized sphere's uniform interior and dipolar exterior as canonical results

1908

— Mie's scattering theory embeds the quasi-static polarized sphere as the small-particle limit — Rayleigh's blue sky quantified

The physical picture

Picture the sphere's surface wearing its bound charge like climate zones: a north polar cap of exposed positive chain-ends, densest at the pole ( $\sigma_b = P$ ), thinning as $\cos\theta$ through the mid-latitudes, vanishing at the equator (where $\mathbf{P}$ skims the surface and no chains terminate), then deepening negative toward the south pole. This is the most important surface charge distribution in the book — it recurs for the conducting sphere in a uniform field (there as $3\varepsilon_0E_0\cos\theta$ , per the separation-of-variables article), for the dielectric sphere, for the magnetized sphere — and its field is worth knowing cold: uniform inside, pure dipole outside.

Figure

The cosine shell — polar caps of bound charge and the two exact fields

Why should a cosine shell produce a uniform interior field? The deep answer is the displaced-spheres picture, which deserves to be seen before any algebra. A uniformly polarized sphere is, microscopically, all its positive charge displaced a whisker $\mathbf{d}$ relative to all its negative charge — as if two rigid balls of charge, $+\rho$ and $-\rho$ , sat almost exactly on top of each other. Where the balls overlap (almost everywhere), the charge cancels: neutral bulk. The tiny crescent of unbalanced charge at the surface — positive where the $+$ ball pokes out, negative opposite, with thickness $d\cos\theta$ — is the cosine shell: $\sigma_b = \rho d\cos\theta = P\cos\theta$ . And now the interior field costs nothing: inside a uniform ball, the field is linear in position (the Gauss workhorse $\mathbf{E} = \rho\mathbf{r}/3\varepsilon_0$ ), so the two balls' interior fields — linear in $\mathbf{r}$ and $\mathbf{r} + \mathbf{d}$ respectively — subtract to a constant. Uniformity is not a coincidence of spheres; it is the fingerprint of interior linearity, the same property that made the crude-model atom a perfect spring two articles back.

Figure

The displaced-spheres construction — two linear fields subtract to a constant

The exterior miracle rides the same construction: outside itself, each uniform ball is a point charge at its own center (the shell theorem), so the exterior field is exactly that of two point charges $\pm q$ a distance $d$ apart — a physical dipole, which in the $d \to 0$ limit at fixed $\rho d$ becomes a pure point dipole with $\mathbf{p} = q\mathbf{d} = \tfrac43\pi R^3\mathbf{P}$ . No quadrupole, no corrections: the multipole ladder truncates after one rung because the sources truncate after one harmonic. That is the Legendre reading, and it is the second derivation's whole content: $\cos\theta$ is $P_1$ , the boundary-value problem responds rung by rung, and a one-rung source excites a one-rung response — uniform field inside ( $r^1$ solution), dipole outside ( $r^{-2}$ solution). Two languages, one truth: the sphere is the shape whose geometry doesn't mix the ladder.

The result

For a sphere of radius $R$ with uniform $\mathbf{P}$ (take $\mathbf{P} = P\hat{\mathbf{z}}$ ), the sources are

\rho_b = -\nabla\cdot\mathbf{P} = 0, \qquad \sigma_b = \mathbf{P}\cdot\hat{\mathbf{r}} = P\cos\theta \qquad (1)

where: $\theta$ — polar angle from the $\mathbf{P}$ direction. The fields:

\mathbf{E}_{\rm in} = -\frac{\mathbf{P}}{3\varepsilon_0} \;\;(\text{uniform, } r < R), \qquad V_{\rm out} = \frac{1}{4\pi\varepsilon_0}\,\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2}, \quad \mathbf{p} = \tfrac43\pi R^3\,\mathbf{P} \;\;(r \geq R) \qquad (2)

where: the exterior potential is the exact point-dipole potential (field pattern per Electric Dipole Moment: $E_r = 2p\cos\theta/4\pi\varepsilon_0r^3$ , $E_\theta = p\sin\theta/4\pi\varepsilon_0r^3$ ), and $\mathbf{p}$ is just polarization times volume — the census reading of $\mathbf{P}$ collapsing the whole sphere to one moment. At the surface the two descriptions meet with the right credentials: the interior potential $V_{\rm in} = (P/3\varepsilon_0)\,r\cos\theta$ and the exterior $V_{\rm out} = (PR^3/3\varepsilon_0)\cos\theta/r^2$ agree at $r = R$ , and the field discontinuities reproduce the E1 boundary conditions on cue (worked example 2 audits them numerically).

Derivation ARoute one — two displaced balls of charge

The goal is both results of (2) in the most economical honest way, and — more valuable than the results — the reason for them. The strategy is the superposition model from the physical picture: a ball of uniform charge $+\rho$ centered at $+\mathbf{d}/2$ and a ball of $-\rho$ centered at $-\mathbf{d}/2$ , in the limit $d \to 0$ with $\mathbf{P} = \rho\,\mathbf{d}$ held fixed. Everything rests on two E1 theorems recalled by name from the Gauss symmetry catalog: inside a uniform ball, $\mathbf{E} = \rho\,\mathbf{r}_c/3\varepsilon_0$ ( $\mathbf{r}_c$ measured from that ball's center); outside, the ball is a point charge at its center. Watch how far the results hold before the limit — the exterior is exact for finite $d$ too, which is what makes the construction more than a trick.

Step 1. The model equals the problem. Overlapping the balls cancels charge wherever both are present; the residue is a surface layer of local thickness $d\cos\theta$ (the gap between two spheres offset along $z$ ), carrying charge per area $\rho\,d\cos\theta = P\cos\theta$ — exactly the bound charge (1) of the uniformly polarized sphere. Same sources, same fields, everywhere: the two problems are one problem.

Step 2. Interior. At field point $\mathbf{r}$ (from the midpoint), the positive ball contributes $\rho(\mathbf{r} - \mathbf{d}/2)/3\varepsilon_0$ , the negative ball $-\rho(\mathbf{r} + \mathbf{d}/2)/3\varepsilon_0$ . Add:

\mathbf{E}_{\rm in} = \frac{\rho}{3\varepsilon_0}\left[(\mathbf{r} - \tfrac{\mathbf{d}}{2}) - (\mathbf{r} + \tfrac{\mathbf{d}}{2})\right] = -\frac{\rho\,\mathbf{d}}{3\varepsilon_0} = -\frac{\mathbf{P}}{3\varepsilon_0}.

The position $\mathbf{r}$ cancels identically: the interior field is uniform, exactly, at any finite $d$ (within the overlap region) — no limit needed. Why: linear functions differ by a constant when offset; interior linearity of the ball field is the entire mechanism, and it is why only ellipsoids (whose interior fields are linear in position for uniform charge) polarize uniformly.

Step 3. Exterior. Outside both balls, each acts as a point charge $\pm q = \pm\tfrac43\pi R^3\rho$ at its center: the exterior field is exactly a physical dipole of separation $d$ . Take $d \to 0$ at fixed $\rho d$ : a pure point dipole with

\mathbf{p} = q\,\mathbf{d} = \tfrac43\pi R^3\rho\,\mathbf{d} = \tfrac43\pi R^3\,\mathbf{P},

and the limit is unnecessary for the moment (it is $q\mathbf{d}$ at any $d$ ) — only for killing the higher multipoles of the physical dipole, which vanish as $O(d^2)$ . In the polarized-sphere reading, $d$ is molecular-displacement scale ( $\sim$ femtometers for real materials, per the induced-dipoles article): the higher multipoles are dead to every conceivable measurement, and in the ideal continuum limit, dead exactly.

Step 4. Consistency sweep. (i) The average-field theorem of the previous article demands $\mathbf{E}_{\rm ave} = -\mathbf{p}/4\pi\varepsilon_0R^3 = -\tfrac43\pi R^3\mathbf{P}/4\pi\varepsilon_0R^3 = -\mathbf{P}/3\varepsilon_0$ — and our interior field, being uniform, is its own average: ✓ (three articles now agree on the $\tfrac13$ ). (ii) The depolarizing hierarchy: $N_{\rm sphere} = \tfrac13$ sits between needle ( $\approx0$ ) and slab ( $1$ ), as the solid-angle intuition predicted. (iii) Total bound charge: $\oint P\cos\theta\,\mathrm{d}a = 0$ by symmetry ✓. Verified: Griffiths Ex. 4.2/4.3 and Prob. 4.11 (the displaced-spheres construction); interior-ball linearity per GAUSA; the moment equals polarization × volume as the census definition requires; the E_ave cross-check closes with FPOL's theorem.

Go deeper · Derivation B · 2 minRoute two — the cosine shell as a one-rung boundary-value problem

Derivation BRoute two — the cosine shell as a one-rung boundary-value problem

The goal is the same pair of results by the Legendre machinery, because this is the route that generalizes — to the conducting sphere, the dielectric sphere, and every axisymmetric shell source — and because it explains the miracle in the ladder's language: a $P_1$ source excites only the $P_1$ response. The strategy is the standard two-region construction from Separation of Variables in Spherical Coordinates, whose general solution $V = \sum_l\left(A_lr^l + B_lr^{-(l+1)}\right)P_l(\cos\theta)$ we recall by name rather than re-derive; the new content is the matching across a charged shell.

Step 1. Set up the two regions. With no charge except the shell at $r = R$ : inside, regularity at the origin kills the $B_l$ terms, $V_{\rm in} = \sum A_lr^lP_l$ ; outside, decay at infinity kills the $A_l$ terms, $V_{\rm out} = \sum B_lr^{-(l+1)}P_l$ .

Step 2. Matching conditions at $r = R$ , imported from the E1 boundary-conditions article: $V$ continuous (no dipole layer), and the normal derivative jumps by the surface charge:

V_{\rm in}(R) = V_{\rm out}(R), \qquad -\left.\frac{\partial V_{\rm out}}{\partial r}\right|_R + \left.\frac{\partial V_{\rm in}}{\partial r}\right|_R = \frac{\sigma_b(\theta)}{\varepsilon_0} = \frac{P\cos\theta}{\varepsilon_0}.

Step 3. Match rung by rung. Legendre orthogonality (the Fourier's-trick of the Legendre article) makes each $l$ independent. The source has only $l = 1$ content — $\cos\theta = P_1(\cos\theta)$ , one rung, exactly — so every $l \neq 1$ coefficient vanishes on the spot, and for $l = 1$ : continuity gives $A_1R = B_1/R^2$ ; the jump condition gives $2B_1/R^3 + A_1 = P/\varepsilon_0$ . Solve the pair:

A_1 = \frac{P}{3\varepsilon_0}, \qquad B_1 = \frac{P R^3}{3\varepsilon_0}.

Step 4. Read off the physics:

V_{\rm in} = \frac{P}{3\varepsilon_0}\,r\cos\theta = \frac{P z}{3\varepsilon_0} \;\Rightarrow\; \mathbf{E}_{\rm in} = -\nabla V_{\rm in} = -\frac{\mathbf{P}}{3\varepsilon_0}; \qquad V_{\rm out} = \frac{PR^3}{3\varepsilon_0}\,\frac{\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}

with $p = 4\pi\varepsilon_0\cdot PR^3/3\varepsilon_0 = \tfrac43\pi R^3P$ — both results of (2), recovered. Why this route is the general-purpose one: any axisymmetric surface density expands as $\sigma(\theta) = \sum\sigma_lP_l(\cos\theta)$ , and each rung solves independently by exactly these two lines of algebra. The uniformly polarized sphere is the cleanest case — a source that is already a single rung — but the identical machinery with an added uniform-field term at infinity solves the conducting sphere (done in the separation-of-variables article) and the dielectric sphere in a field (done in its own article, two ahead). One method, whole family. Verified: coefficients checked by back-substitution into both matching conditions; V continuous at r = R ✓; the l = 1 jump reproduces P cosθ/ε0 ✓; agreement with Route one is exact, and the two derivations share no steps — independent confirmations.

Limiting cases

Case	Result	Meaning
interior, anywhere	$\mathbf{E} = -\mathbf{P}/3\varepsilon_0$ , uniform	the exact $\tfrac13$ ; only ellipsoids manage this
exterior, any $r \geq R$	pure dipole, $\mathbf{p} = \tfrac43\pi R^3\mathbf{P}$	ladder truncates: one-harmonic source
far field	$V \to p\cos\theta/4\pi\varepsilon_0r^2$	consistent with any neutral object's leading term
surface, equator	$\sigma_b = 0$ , $E_\theta$ continuous $= P\sin\theta/3\varepsilon_0\cdot$ ... at $\theta = 90°$ : $P/3\varepsilon_0$	tangential continuity with no charge
surface, pole	normal jump $E_r^{\rm out} - E_r^{\rm in} = P/\varepsilon_0$	$\sigma_b(0)/\varepsilon_0$ : BCEV verified
needle / slab comparison	$N$ : $0$ / $\tfrac13$ / $1$	sphere is the exact middle of the hierarchy
conducting sphere in field $E_0$	$\sigma = 3\varepsilon_0E_0\cos\theta$	same cosine shell, run backwards (SoV article)

Common trapThe interior field is not the field of the dipole formula continued inward

The exterior is a pure dipole field, and the temptation is to extend $E \propto (2\cos\theta\,\hat{\mathbf{r}} + \sin\theta\,\hat{\boldsymbol{\theta}})/r^3$ inside the sphere — where it would diverge at the center and point the wrong way at the equator. The truth switches character at $r = R$ : outside, dipolar and falling as $1/r^3$ ; inside, constant $-\mathbf{P}/3\varepsilon_0$ . (Field lines make the point vividly: outside they arc pole to pole; inside they run straight and parallel, south-cap to north-cap, against $\mathbf{P}$ .) A related trap: the interior field opposes $\mathbf{P}$ — a frozen electret sphere's own field works to depolarize it — but do not conclude the polarization "cancels itself"; for a permanent (frozen) $\mathbf{P}$ the depolarizing field just sits there, and for a linear dielectric the self-consistent response settles at a finite reduced value, computed via exactly this $\tfrac13$ in the sphere-in-a-field article.

Worked examples

1 · An electret sphere, fully evaluated (confidence builder)

A polymer electret sphere, $R = 2.0\,\mathrm{cm}$ , carries frozen uniform $P = 1.0\times10^{-4}\,\mathrm{C/m^2}$ . Find the surface charge extremes, the interior field, the dipole moment, and the potential at the north pole of the surface.

Step 1. Surface charge: $\sigma_b = P\cos\theta$ — maximum $+1.0\times10^{-4}\,\mathrm{C/m^2}$ at the north pole, zero at the equator, $-1.0\times10^{-4}$ at the south pole. ( $\rho_b = 0$ : uniform $\mathbf{P}$ .)

Step 2. Interior field: $E_{\rm in} = P/3\varepsilon_0 = 10^{-4}/(3\times8.854\times10^{-12}) = 3.8\times10^{6}\,\mathrm{V/m}$ , pointing from the $+$ cap toward the $-$ cap — antiparallel to $\mathbf{P}$ , uniform to the last interior point. (Note the scale: an internal field above air's breakdown, from modest frozen polarization — electret fields are fierce.)

Step 3. Dipole moment: $p = \tfrac43\pi R^3P = \tfrac43\pi(0.02)^3(10^{-4}) = 3.4\times10^{-9}\,\mathrm{C\cdot m}$ .

Step 4. Pole potential: $V(R, 0) = PR/3\varepsilon_0 = (10^{-4})(0.02)/(3\times8.854\times10^{-12}) = 7.5\times10^{4}\,\mathrm{V}$ — seventy-five kilovolts on a golf-ball-sized wax sphere, with total charge exactly zero. On the axis at $r = 10\,\mathrm{cm}$ , the exterior dipole formula gives $E = 2p/4\pi\varepsilon_0r^3 = 6.0\times10^{4}\,\mathrm{V/m}$ — the whole sphere, impersonating a single point dipole, exactly.

2 · Auditing the boundary — the E1 matching conditions on a live example (exam level)

Verify, with the sphere of example 1, that the fields of equation (2) satisfy the electrostatic boundary conditions at the surface: tangential $\mathbf{E}$ continuous, normal $\mathbf{E}$ jumping by $\sigma_b/\varepsilon_0$ — at both the pole and the equator.

Step 1. Gather the surface fields. Inside (uniform): $\mathbf{E}_{\rm in} = -(P/3\varepsilon_0)\hat{\mathbf{z}}$ ; in spherical components, $E_r^{\rm in} = -(P/3\varepsilon_0)\cos\theta$ , $E_\theta^{\rm in} = +(P/3\varepsilon_0)\sin\theta$ . Outside at $r = R$ (dipole with $p/4\pi\varepsilon_0R^3 = P/3\varepsilon_0$ ): $E_r^{\rm out} = (2P/3\varepsilon_0)\cos\theta$ , $E_\theta^{\rm out} = (P/3\varepsilon_0)\sin\theta$ .

Step 2. Tangential continuity, everywhere: $E_\theta^{\rm out} - E_\theta^{\rm in} = (P/3\varepsilon_0)\sin\theta - (P/3\varepsilon_0)\sin\theta = 0$ ✓ at every $\theta$ — as it must be, since $E_\parallel$ continuity is charge-blind (it comes from $\nabla\times\mathbf{E} = 0$ , not from Gauss).

Step 3. Normal jump: $E_r^{\rm out} - E_r^{\rm in} = (2P/3\varepsilon_0)\cos\theta - (-(P/3\varepsilon_0)\cos\theta) = (P/\varepsilon_0)\cos\theta = \sigma_b(\theta)/\varepsilon_0$ ✓ at every $\theta$ . Numbers at the pole: outside $7.5\times10^{6}$ , inside $-3.8\times10^{6}$ , jump $1.13\times10^{7}\,\mathrm{V/m} = \sigma_b(0)/\varepsilon_0$ with $\sigma_b(0) = 10^{-4}\,\mathrm{C/m^2}$ ✓. At the equator: $\sigma_b = 0$ and indeed both components continuous ( $E_r = 0$ both sides; $E_\theta = P/3\varepsilon_0 = 3.8\times10^{6}$ both sides).

Step 4. Read what the audit teaches. The boundary conditions are not extra physics to impose — they are properties the correct solution exhibits. Here they also decode the geometry: the field pierces the surface radially at the poles (max charge, max jump) and slides along it at the equator (no charge, no jump) — the cosine shell's fingerprint, readable straight off the matching conditions.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §4.2 (Ex. 4.2–4.3, Prob. 4.11) · Jackson, Classical Electrodynamics, 3rd ed., §4.3 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §10.9 · Zangwill, Modern Electrodynamics, §6.4.

Provenance. The uniformly polarized sphere results $\mathbf{E}_{\rm in}=-\mathbf{P}/3\varepsilon_0$ and exterior pure-dipole potential with $\mathbf{p}=\tfrac43\pi R^3\mathbf{P}$ are standard (Griffiths Ex. 4.2–4.3, Jackson §4.3); the displaced-spheres construction is Griffiths Prob. 4.11 register; the Legendre matching uses the separation-of-variables machinery (Griffiths §3.3) with the surface-charge jump condition. Constant $\varepsilon_0=8.854\times10^{-12}$ F/m. A verifier should rerun both routes and the boundary-condition audit. Worked numbers recomputed for this edition: $E_{\rm in} = 3.8\times10^6$ V/m, $p = 3.4\times10^{-9}$ C·m, pole potential $7.5\times10^4$ V, on-axis exterior field at 10 cm $= 6.0\times10^4$ V/m; boundary audit — pole jump $1.13\times10^7$ V/m $= \sigma_b(0)/\varepsilon_0$ , equatorial $E_\theta = 3.8\times10^6$ V/m continuous.

Builds on: Bound Charges: Surface and Volume · Field of a Polarized Object · Separation of Variables in Spherical Coordinates · Legendre Polynomials · Boundary Conditions on E and V

Leads to: The Electric Displacement D · Dielectric Sphere in a Uniform Field · Local Field and the Lorentz Cavity · Clausius–Mossotti Relation

Is the interior field really uniform all the way to the surface, or only near the center?

All the way — that is the astonishing part. Both derivations deliver exact uniformity for every interior point: the displaced-balls route because linear fields subtract to constants throughout the overlap, the Legendre route because $V_{\rm in} = A_1r\cos\theta = A_1z$ is globally linear in $z$ . A millimeter from the north pole, where the positive cap looms, the field is the same $-P/3\varepsilon_0$ as at the center: the nearby cap's extra push is exactly compensated by the far cap's pull and the mid-latitudes' geometry. Perturb the shape even slightly (an egg, a dented sphere) and the exact uniformity is gone — it survives only for ellipsoids, with the $\tfrac13$ splitting into three axis-dependent factors summing to 1.

Why does the factor 1/3 keep appearing all over this chapter?

Because the sphere is the default shape of isotropy, and $\tfrac13$ is the sphere's share of the depolarizing budget. The three ellipsoid depolarizing factors along principal axes always sum to 1 (a trace identity); isotropy forces them equal, so each is $\tfrac13$ . The same number then surfaces wherever a spherical region is singled out: the average-field theorem ( $-\mathbf{p}/4\pi\varepsilon_0R^3$ is $-\mathbf{P}/3\varepsilon_0$ in disguise), the Lorentz cavity field ( $+\mathbf{P}/3\varepsilon_0$ , the hole's version), the Clausius–Mossotti denominator, the plasmon resonance condition $\varepsilon = -2\varepsilon_0$ . One geometric fact — a sphere's cosine shell makes a uniform field of strength one third — wearing five costumes.

If I place the electret sphere in an external field, do the results here still help?

Directly — by superposition. A frozen (permanent) $\mathbf{P}$ doesn't respond to the field, so the total is just external field plus this article's field: uniform $\mathbf{E}_0 - \mathbf{P}/3\varepsilon_0$ inside, external-plus-dipole outside; the sphere also feels the torque and gradient forces of the dipole-mechanics article via its total moment $\mathbf{p} = \tfrac43\pi R^3\mathbf{P}$ . For a linear dielectric sphere the polarization itself is field-induced and must be found self-consistently — $\mathbf{P}$ responds to a field that includes its own $-\mathbf{P}/3\varepsilon_0$ — a two-line fixed-point calculation that the Dielectric Sphere in a Uniform Field article runs; the $\tfrac13$ from here is its key input.

Where does the energy of the electret sphere reside, and how much is there?

In the field, with density $\tfrac{\varepsilon_0}{2}E^2$ (per the E1 field-energy article). The interior contributes $\tfrac{\varepsilon_0}{2}(P/3\varepsilon_0)^2\cdot\tfrac43\pi R^3 = \tfrac{2\pi R^3P^2}{27\varepsilon_0}$ ; the exterior dipole field contributes exactly twice that (a classic integral), for a total $W = 2\pi R^3P^2/9\varepsilon_0$ — about $0.7\,\mathrm{mJ}$ for our example sphere. Note the sign of the physics: this positive field energy is the work that had to be done against the depolarizing field to assemble the frozen polarization, and it is why electrets are metastable — the system would happily shed that energy if the molecular locking ever released.

The conducting sphere in a uniform field had sigma proportional to cos theta too — same physics?

Same mathematics, mirrored motivation. The conductor in field $E_0$ develops $\sigma = 3\varepsilon_0E_0\cos\theta$ , precisely the cosine shell that makes a uniform interior field — the conductor tunes its shell until the interior total vanishes, which is what "conductor" means. Solve $P/3\varepsilon_0$ -style bookkeeping backwards and you recover its induced moment $p = 4\pi\varepsilon_0R^3E_0$ (the polarizability from the images article). The electret runs the same shell at fixed strength with no cancellation demand; the linear dielectric sits between, cancelling partially. Three problems, one cosine shell, distinguished only by what determines its amplitude.

Do the two derivation routes ever disagree — say, for a uniformly polarized cube?

They can't disagree where both apply; the instructive point is that for a cube neither yields closed uniformity — because there is none. The displaced-cubes construction still gives the right sources (face charges), but a cube's interior field for uniform charge is not linear in position, so the subtraction is not constant: the interior field of a polarized cube is genuinely non-uniform (strongest near face centers, weaker toward edges), computable only numerically or by series. The Legendre route needs spherical boundaries and fails structurally. Moral: the sphere's miracles are earned by its geometry, and the general polarized object keeps only the guarantees of the equivalence theorem — correct sources, correct fields, no promise of prettiness.

How is this construction used for magnets — is a uniformly magnetized sphere really the same problem?

Formally identical in the potential language: a uniform magnetization $\mathbf{M}$ has "bound magnetic charge" $\sigma_m = \mathbf{M}\cdot\hat{\mathbf{n}} = M\cos\theta$ in the scalar-potential formulation, so the $\mathbf{H}$ -field is uniform inside ( $-\mathbf{M}/3$ ) and dipolar outside — the same $\tfrac13$ , now called the demagnetizing factor. The physical field $\mathbf{B} = \mu_0(\mathbf{H} + \mathbf{M})$ inside then comes out along $\mathbf{M}$ at $+\tfrac23\mu_0\mathbf{M}$ , a sign flip with real consequences (compass needles, MRI shimming). The magnetostatics chapter (E5) rebuilds this honestly with bound currents; the sphere solved here is the Rosetta stone between the two descriptions.