Bound Charge and the Displacement Field

Derivation

Bound Charge and the Displacement Field

D-287 Home PU-305 Threads fields · matter Depends on Multipole Expansion of the Potential, Gauss's Law from Coulomb's Law

Statement

The macroscopic electrostatic potential of a body carrying polarization \(\mathbf{P}(\mathbf{r})\) (electric dipole moment per unit volume) is exactly the potential produced by a bound volume charge \(\rho_b = -\nabla\cdot\mathbf{P}\) together with a bound surface charge \(\sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}}\). Feeding these bound charges into Gauss's law and separating them from the free charge \(\rho_f\) yields the auxiliary displacement field \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) obeying \(\nabla\cdot\mathbf{D}=\rho_f\).

Why it matters

Matter contains an astronomical number of atomic charges; solving Coulomb's law for each is hopeless. Coarse-graining those charges into a smooth dipole density \(\mathbf{P}\) and proving that only its divergence and boundary discontinuity act as sources reduces the whole problem to two effective charge densities. This is the bridge from microscopic electrostatics to the macroscopic Maxwell equations in media.

The field \(\mathbf{D}\) is engineered so that its Gauss law "sees" only the charge we control — the free charge on capacitor plates, injected carriers, external ions — while the response of the medium is bookkept inside \(\mathbf{P}\). Every treatment of dielectrics, capacitance with insulators, and boundary conditions at material interfaces rests on this result.

Assumptions

The medium is describable by a well-defined dipole moment per unit volume \(\mathbf{P}(\mathbf{r})\).If the coarse-graining volume is too small to contain many atoms, \(\mathbf{P}\) is not smooth and \(\rho_b=-\nabla\cdot\mathbf{P}\) is meaningless; the field must be computed charge-by-charge.

Each macroscopic volume element is charge-neutral apart from its dipole moment (no intrinsic monopole term, no free charge counted here).If a volume element carries net free charge, that contribution belongs to \(\rho_f\) and must be added separately; \(\rho_b\) alone would then misstate the total source.

Higher multipoles (quadrupole and beyond) of each volume element are negligible at the macroscopic scale.If retained, additional effective sources such as \(\nabla\nabla{:}\mathbf{Q}\) appear; \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) acquires quadrupole corrections and the clean form is lost. This is subtle because such terms are usually tiny but nonzero in ordered media.

Static (or quasi-static) fields, so \(\nabla\times\mathbf{E}=0\) and a scalar potential exists.If time-dependent, \(\partial\mathbf{P}/\partial t\) contributes a polarization current \(\mathbf{J}_b=\partial\mathbf{P}/\partial t\) and the story extends to the full Maxwell equations; the electrostatic derivation below no longer captures everything.

Derivation

\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal V} \frac{\mathbf{P}(\mathbf{r}')\cdot(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^{3}}\, d^3 r' \]

Each volume element \(d^3r'\) carries dipole moment \(d\mathbf{p}=\mathbf{P}\,d^3r'\); superpose the dipole potential from the multipole-expansion result. A

\[ \nabla'\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) = \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} \]

Identity for the gradient of the Coulomb kernel with respect to the source coordinate \(\mathbf{r}'\); rewrites the awkward vector kernel as a gradient. A

\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal V} \mathbf{P}(\mathbf{r}')\cdot\nabla'\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) d^3 r' \]

Substitute the identity of Step 2 into Step 1. A

\[ \mathbf{P}\cdot\nabla'\!\left(\frac{1}{\mathfrak{r}}\right) = \nabla'\!\cdot\!\left(\frac{\mathbf{P}}{\mathfrak{r}}\right) - \frac{1}{\mathfrak{r}}\,\nabla'\!\cdot\mathbf{P}, \qquad \mathfrak{r}\equiv|\mathbf{r}-\mathbf{r}'| \]

Product rule for the divergence, \(\nabla\cdot(f\mathbf{A})=\mathbf{A}\cdot\nabla f + f\,\nabla\cdot\mathbf{A}\), rearranged to isolate \(\mathbf{P}\cdot\nabla'(1/\mathfrak r)\). This is the pivotal algebraic move. B

\[ V = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal V}\nabla'\!\cdot\!\left(\frac{\mathbf{P}}{\mathfrak r}\right)d^3r' \;-\; \frac{1}{4\pi\varepsilon_0}\int_{\mathcal V}\frac{\nabla'\!\cdot\mathbf{P}}{\mathfrak r}\,d^3r' \]

Insert Step 4 and split the integral by linearity. A

\[ \int_{\mathcal V}\nabla'\!\cdot\!\left(\frac{\mathbf{P}}{\mathfrak r}\right)d^3r' = \oint_{\partial\mathcal V}\frac{\mathbf{P}\cdot\hat{\mathbf{n}}}{\mathfrak r}\,da' \]

Divergence theorem converts the first volume integral to a surface integral over the boundary \(\partial\mathcal V\) with outward normal \(\hat{\mathbf n}\). B

\[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\oint_{\partial\mathcal V}\frac{\mathbf{P}\cdot\hat{\mathbf{n}}}{\mathfrak r}\,da' \;+\; \frac{1}{4\pi\varepsilon_0}\int_{\mathcal V}\frac{(-\nabla'\!\cdot\mathbf{P})}{\mathfrak r}\,d^3r' \]

Combine Steps 5 and 6. Structurally this is exactly a surface charge \(\sigma_b\) plus a volume charge \(\rho_b\). A

\[ \boxed{\;\sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}}, \qquad \rho_b = -\nabla\cdot\mathbf{P}\;} \]

Match Step 7 term-by-term to the standard potential \(V=\frac{1}{4\pi\varepsilon_0}\big(\oint \sigma_b/\mathfrak r\,da' + \int \rho_b/\mathfrak r\,d^3r'\big)\); identical kernels force identical sources. A

\[ \nabla\cdot\mathbf{E} = \frac{\rho_{\text{tot}}}{\varepsilon_0} = \frac{\rho_f + \rho_b}{\varepsilon_0} = \frac{\rho_f - \nabla\cdot\mathbf{P}}{\varepsilon_0} \]

Gauss's law in differential form (from Coulomb's law) applied to the total charge, then substitute \(\rho_b\) from Step 8. A

\[ \nabla\cdot(\varepsilon_0\mathbf{E}+\mathbf{P}) = \rho_f \]

Move \(\nabla\cdot\mathbf{P}\) to the left and factor the divergence (linear operator). The natural combination \(\varepsilon_0\mathbf{E}+\mathbf{P}\) emerges. A

\[ \mathbf{D}\equiv\varepsilon_0\mathbf{E}+\mathbf{P} \quad\Longrightarrow\quad \nabla\cdot\mathbf{D}=\rho_f \]

Define the displacement field as that combination; its Gauss law is sourced by free charge alone. A

Result

\[ \rho_b = -\nabla\cdot\mathbf{P},\qquad \sigma_b=\mathbf{P}\cdot\hat{\mathbf{n}},\qquad \mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P},\qquad \nabla\cdot\mathbf{D}=\rho_f \]

Reading. A region where the polarization "piles up" — where more dipole heads point in than tails point out, i.e. \(\nabla\cdot\mathbf{P}>0\) — hosts a net negative bound charge, because dipole tails (negative ends) accumulate there. At a surface the outward-pointing component \(\mathbf{P}\cdot\hat{\mathbf n}\) leaves exposed positive ends. The displacement \(\mathbf{D}\) repackages Gauss's law so its only source is the charge you actually deposit, \(\rho_f\); the medium's reaction is hidden inside \(\mathbf{P}\).

Units check. \([\mathbf{P}]=\text{C·m·m}^{-3}=\text{C·m}^{-2}\). Then \([\nabla\cdot\mathbf{P}]=\text{C·m}^{-2}/\text{m}=\text{C·m}^{-3}\), a volume charge density — consistent with \(\rho_b\). And \([\mathbf{P}\cdot\hat{\mathbf n}]=\text{C·m}^{-2}\), a surface charge density — consistent with \(\sigma_b\). Also \([\varepsilon_0\mathbf{E}]=(\text{C}^2\text{N}^{-1}\text{m}^{-2})(\text{N·C}^{-1})=\text{C·m}^{-2}=[\mathbf{P}]\), so \(\mathbf{D}\) is uniform in units, and \([\nabla\cdot\mathbf{D}]=\text{C·m}^{-3}=[\rho_f]\).

Limiting cases

Uniform polarization (\(\mathbf{P}=\text{const}\)): \(\nabla\cdot\mathbf{P}=0\), so \(\rho_b=0\) everywhere inside; all bound charge lives on the surface as \(\sigma_b=\mathbf{P}\cdot\hat{\mathbf n}\) (e.g. a uniformly polarized sphere or bar electret).
Vacuum (\(\mathbf{P}=0\)): \(\mathbf{D}=\varepsilon_0\mathbf{E}\) and \(\nabla\cdot\mathbf{D}=\rho_f\) collapses to ordinary Gauss's law.
Linear isotropic dielectric (\(\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}\)): \(\mathbf{D}=\varepsilon_0(1+\chi_e)\mathbf{E}=\varepsilon\mathbf{E}\), recovering the constitutive relation with permittivity \(\varepsilon=\varepsilon_0\varepsilon_r\).
No free charge (\(\rho_f=0\)): \(\nabla\cdot\mathbf{D}=0\); \(\mathbf{D}\) is source-free even though \(\mathbf{E}\) is not, since \(\nabla\cdot\mathbf{E}=\rho_b/\varepsilon_0\neq0\).

Breaks when

Sharp material boundaries / point-like sampling. The identification \(\rho_b=-\nabla\cdot\mathbf{P}\) assumes \(\mathbf{P}\) varies smoothly. Right at an atomically sharp edge, or if you probe on the scale of the interatomic spacing, \(\mathbf{P}\) is discontinuous and the divergence becomes a surface delta — the volume formula alone misses the surface term \(\sigma_b\). One must always carry the boundary contribution.
Strong or high-order multipole media. If the microscopic charge distribution of a cell has a significant quadrupole moment (some ordered crystals, dense strongly-correlated matter), the pure-dipole coarse-graining fails; effective sources beyond \(-\nabla\cdot\mathbf{P}\) appear and \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) is only approximate.
Rapid time dependence. Under fast driving, \(\partial\mathbf{P}/\partial t\) is a real current \(\mathbf{J}_b\); the electrostatic potential picture breaks and one needs the full Maxwell–Ampère law. \(\mathbf{D}\) still exists but its dynamics couple to \(\mathbf{B}\).
Nonlocal / spatially dispersive response. If \(\mathbf{P}(\mathbf{r})\) depends on \(\mathbf{E}\) at neighbouring points (nonlocal \(\chi_e\)), no local \(\rho_b(\mathbf{r})\) captures the response and the simple constitutive closure is invalid.

Failure modes

Sign flip on \(\rho_b\). Writing \(\rho_b=+\nabla\cdot\mathbf{P}\). The minus sign is physical: divergence of \(\mathbf{P}\) means dipole positive ends fan outward, leaving the negative tails behind.
Dropping the surface charge. Computing only \(\rho_b=-\nabla\cdot\mathbf{P}\) for a uniformly polarized object and concluding there is no field — forgetting \(\sigma_b=\mathbf{P}\cdot\hat{\mathbf n}\), which is the only source there.
Treating \(\mathbf{D}\) as "the field in the dielectric." The force on a test charge is \(q\mathbf{E}\), never \(q\mathbf{D}\); \(\mathbf{D}\) is an auxiliary bookkeeping field.
Assuming \(\nabla\times\mathbf{D}=0\). Only \(\nabla\times\mathbf{E}=0\) in statics. Since \(\nabla\times\mathbf{D}=\nabla\times\mathbf{P}\), \(\mathbf{D}\) can be non-conservative; you cannot define a scalar "\(\mathbf{D}\)-potential" in general.
Using \(\mathbf{D}=\varepsilon\mathbf{E}\) universally. That constitutive relation holds only for linear isotropic media; the definition \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) is always valid but \(\mathbf{P}\) need not be proportional to \(\mathbf{E}\).
Confusing bound and free charge in Gauss's law. Enclosing bound charge in a Gaussian surface for \(\mathbf{D}\): \(\oint\mathbf{D}\cdot d\mathbf{a}=Q_{f,\text{enc}}\) counts free charge only.

Discussion

The heart of the derivation is Step 4: a single application of the product rule converts the microscopic-looking dipole integral into two terms with the unmistakable structure of Coulomb potentials from a volume charge and a surface charge. Nothing is approximated in Steps 1–8 — given a smooth \(\mathbf{P}\), the equivalence to \(\rho_b\) and \(\sigma_b\) is exact. The bound charge is not a mathematical fiction: it is genuine, measurable charge (an electret deflects an electroscope), arising from the incomplete cancellation of aligned atomic dipoles. Where the dipoles are uniform, interior charges cancel head-to-tail and only the unpaired ends at the surface survive; where \(\mathbf{P}\) has a gradient, the cancellation is imperfect in the bulk and \(-\nabla\cdot\mathbf{P}\) remains.

The passage from Step 9 to Step 11 is deliberately unglamorous but conceptually decisive. Total charge splits as \(\rho_{\text{tot}}=\rho_f+\rho_b\); the \(\rho_b\) piece is itself \(-\nabla\cdot\mathbf{P}\), a divergence, so it can be absorbed into the divergence operator on the left. What is left, \(\nabla\cdot(\varepsilon_0\mathbf{E}+\mathbf{P})=\rho_f\), isolates exactly the charge an experimenter deposits. \(\mathbf{D}\) is engineered for this convenience and nothing more: it lets us apply Gauss's-law arguments in symmetric geometries (slab, sphere, cylinder) using only the free charge we control.

The connection to the constitutive relation is where physics re-enters. \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) is a definition and carries no information about the material; the material physics lives in the relation \(\mathbf{P}(\mathbf{E})\). For weak fields in ordinary insulators this linearizes to \(\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}\), giving \(\mathbf{D}=\varepsilon\mathbf{E}\) and permittivity \(\varepsilon=\varepsilon_0\varepsilon_r\). Ferroelectrics carry \(\mathbf{P}\neq0\) at \(\mathbf{E}=0\); anisotropic crystals make \(\chi_e\) a tensor so \(\mathbf{D}\) and \(\mathbf{E}\) need not be parallel — the definition survives all of these, which is precisely its value.

At the deepest level the split into "free" and "bound" is a choice of coarse-graining scale, not a law of nature. The exact microscopic field obeys \(\nabla\cdot\mathbf{e}=\rho_{\text{micro}}/\varepsilon_0\) with the full atomic \(\rho_{\text{micro}}\). Averaging over a mesoscopic volume, the dipole term of each cell's multipole expansion produces \(-\nabla\cdot\mathbf{P}\); had we retained the quadrupole term we would find an additional effective density \(+\tfrac12\partial_i\partial_j Q_{ij}\), redefining \(\mathbf{D}\) with a quadrupole correction. That \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\) works so universally is a statement that macroscopic fields are overwhelmingly dominated by the dipole density — a happy but contingent fact about ordinary matter, not a theorem.

Common misconceptions. \(\mathbf{D}\) is not "the electric field inside a dielectric" and does not give the force on charges — that is always \(\mathbf{E}\). Bound charge is real charge, not an accounting trick. And \(\nabla\cdot\mathbf{D}=\rho_f\) does not mean \(\mathbf{D}\) is unaffected by the medium — its value everywhere still depends on the geometry and on \(\mathbf{P}\) through the boundary conditions; only its divergence is blind to bound charge.

Worked examples

Example 1 — Uniformly polarized dielectric sphere.

\[ \mathbf{P}=P_0\,\hat{\mathbf z}\ (\text{const}),\qquad \rho_b=-\nabla\cdot\mathbf{P}=0 \]

Constant vector field has zero divergence, so no interior bound charge. A

\[ \sigma_b=\mathbf{P}\cdot\hat{\mathbf n}=P_0\,\hat{\mathbf z}\cdot\hat{\mathbf r}=P_0\cos\theta \]

On a sphere the outward normal is \(\hat{\mathbf r}\); \(\hat{\mathbf z}\cdot\hat{\mathbf r}=\cos\theta\). A

\[ P_0 = 3.0\times10^{-6}\ \text{C·m}^{-2},\quad \sigma_b(\text{pole})=P_0\cos0=3.0\times10^{-6}\ \text{C·m}^{-2} \]

Insert numbers at the north pole (\(\theta=0\)); the equator (\(\theta=90^\circ\)) gives \(\sigma_b=0\). A

\[ \mathbf{E}_{\text{in}}=-\frac{\mathbf{P}}{3\varepsilon_0}=-\frac{3.0\times10^{-6}}{3(8.85\times10^{-12})}\,\hat{\mathbf z}=-1.13\times10^{5}\ \text{V·m}^{-1}\,\hat{\mathbf z} \]

Standard result for the uniform interior field of a \(\cos\theta\) surface-charge sphere (solved via Legendre expansion); quoted here as it follows from this bound charge. B

\[ \rho_b=0,\quad \sigma_b=P_0\cos\theta,\quad \mathbf{E}_{\text{in}}=-\tfrac{P_0}{3\varepsilon_0}\hat{\mathbf z}\approx-1.1\times10^{5}\ \text{V·m}^{-1}\,\hat{\mathbf z} \]

Reading. All bound charge sits on the surface as a \(\cos\theta\) band — positive cap up, negative cap down — producing a uniform depolarizing field inside that opposes \(\mathbf{P}\).

Units check. \(P_0/\varepsilon_0\) has units \((\text{C·m}^{-2})/(\text{C}^2\text{N}^{-1}\text{m}^{-2})=\text{N·C}^{-1}=\text{V·m}^{-1}\). Correct for a field.

Example 2 — Radially polarized shell and its free-charge Gauss law.

\[ \mathbf{P}(r)=\frac{k}{r}\,\hat{\mathbf r}\quad (a\le r\le b),\qquad k=2.0\times10^{-6}\ \text{C·m}^{-1} \]

Given radial polarization in a spherical shell; \(k\) chosen so \([\mathbf P]=\text{C·m}^{-2}\). A

\[ \rho_b=-\nabla\cdot\mathbf{P}=-\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\cdot\frac{k}{r}\right)=-\frac{1}{r^2}\frac{d}{dr}(kr)=-\frac{k}{r^2} \]

Spherical divergence of a radial field, \(\nabla\cdot\mathbf{A}=\frac{1}{r^2}\frac{d}{dr}(r^2A_r)\). B

\[ \sigma_b(a)=\mathbf{P}\cdot\hat{\mathbf n}=\frac{k}{a}(\hat{\mathbf r}\cdot(-\hat{\mathbf r}))=-\frac{k}{a},\qquad \sigma_b(b)=+\frac{k}{b} \]

Inner surface has outward normal \(-\hat{\mathbf r}\); outer surface \(+\hat{\mathbf r}\). A

\[ Q_b^{\text{tot}}=\int_a^b\!\Big(\!-\frac{k}{r^2}\Big)4\pi r^2\,dr + 4\pi a^2\Big(\!-\frac{k}{a}\Big)+4\pi b^2\Big(\frac{k}{b}\Big) = -4\pi k(b-a)-4\pi k a+4\pi k b=0 \]

Sum volume and both surface bound charges; total bound charge of a neutral polarized body must vanish — a check. B

\[ \text{With no free charge: }\oint\mathbf{D}\cdot d\mathbf{a}=Q_{f,\text{enc}}=0\ \Rightarrow\ \mathbf{D}=0,\quad \mathbf{E}=-\mathbf{P}/\varepsilon_0=-\frac{k}{\varepsilon_0 r}\hat{\mathbf r} \]

Spherical symmetry plus \(\rho_f=0\) forces \(\mathbf D=0\); then \(\mathbf E=(\mathbf D-\mathbf P)/\varepsilon_0=-\mathbf P/\varepsilon_0\). B

\[ \rho_b=-\frac{k}{r^2},\quad \sigma_b(a)=-\frac{k}{a},\ \sigma_b(b)=\frac{k}{b},\quad Q_b^{\text{tot}}=0,\quad \mathbf{E}=-\frac{k}{\varepsilon_0 r}\hat{\mathbf r} \]

Reading. Because there is no free charge, \(\mathbf{D}\) vanishes everywhere by symmetry even though \(\mathbf{E}\) does not — a vivid demonstration that \(\mathbf{D}\) tracks only free charge while \(\mathbf{E}\) responds to the bound charge.

Units check. \(k/r^2=(\text{C·m}^{-1})/\text{m}^2=\text{C·m}^{-3}=[\rho_b]\); \(k/a=(\text{C·m}^{-1})/\text{m}=\text{C·m}^{-2}=[\sigma_b]\); \(k/(\varepsilon_0 r)=(\text{C·m}^{-1})/[(\text{C}^2\text{N}^{-1}\text{m}^{-2})\,\text{m}]=\text{N·C}^{-1}=\text{V·m}^{-1}\). All consistent.

Problems

(A) A slab \(0\le z\le d\) has \(\mathbf{P}=P_0\,\hat{\mathbf z}\). Find \(\rho_b\) and the bound surface charges on both faces. Take \(P_0=4.0\times10^{-6}\ \text{C·m}^{-2}\).
Solution
\(\rho_b=-\nabla\cdot\mathbf{P}=0\) (uniform). Top face \(z=d\): \(\hat{\mathbf n}=+\hat{\mathbf z}\), \(\sigma_b=P_0\cdot(+1)=+4.0\times10^{-6}\ \text{C·m}^{-2}\). Bottom face \(z=0\): \(\hat{\mathbf n}=-\hat{\mathbf z}\), \(\sigma_b=P_0\cdot(-1)=-4.0\times10^{-6}\ \text{C·m}^{-2}\). Net bound charge zero, as required.
(A/B) A long cylinder of radius \(a\) has \(\mathbf{P}=\alpha s\,\hat{\mathbf s}\) (cylindrical radius \(s\)), \(\alpha=5.0\times10^{-4}\ \text{C·m}^{-3}\), \(a=0.020\ \text{m}\). Find \(\rho_b\) and \(\sigma_b\).
Solution
\(\rho_b=-\frac{1}{s}\frac{d}{ds}(s\cdot\alpha s)=-\frac{1}{s}\frac{d}{ds}(\alpha s^2)=-\frac{1}{s}(2\alpha s)=-2\alpha=-1.0\times10^{-3}\ \text{C·m}^{-3}\). Surface \(s=a\): \(\sigma_b=P(a)=\alpha a=(5.0\times10^{-4})(0.020)=1.0\times10^{-5}\ \text{C·m}^{-2}\). Check per unit length: volume \(\int_0^a(-2\alpha)2\pi s\,ds=-2\alpha\pi a^2\); surface \(\alpha a\cdot2\pi a=2\alpha\pi a^2\); sum \(=0\). Good.
(B) A parallel-plate capacitor, plate area \(A=0.010\ \text{m}^2\), gap \(d=1.0\ \text{mm}\), free surface charge \(\sigma_f=2.0\times10^{-6}\ \text{C·m}^{-2}\), is filled with a linear dielectric \(\varepsilon_r=4.0\). Find \(\mathbf{D}\), \(\mathbf{E}\), \(\mathbf{P}\), and the bound surface charge on the dielectric.
Solution
\(\mathbf{D}\): Gauss for \(\mathbf D\) gives \(D=\sigma_f=2.0\times10^{-6}\ \text{C·m}^{-2}\). \(E=D/(\varepsilon_0\varepsilon_r)=\frac{2.0\times10^{-6}}{(8.85\times10^{-12})(4.0)}=5.65\times10^{4}\ \text{V·m}^{-1}\). \(P=D-\varepsilon_0 E=2.0\times10^{-6}-(8.85\times10^{-12})(5.65\times10^{4})=2.0\times10^{-6}-5.0\times10^{-7}=1.5\times10^{-6}\ \text{C·m}^{-2}\). Bound surface charge \(\sigma_b=\mathbf P\cdot\hat{\mathbf n}=-P=-1.5\times10^{-6}\ \text{C·m}^{-2}\) on the face adjacent to the positive plate (normal points toward the plate, opposite \(\mathbf P\)). Equivalently \(\sigma_b=-\sigma_f(1-1/\varepsilon_r)=-2.0\times10^{-6}(0.75)=-1.5\times10^{-6}\ \text{C·m}^{-2}\).
(B) A sphere of radius \(R\) carries \(\mathbf{P}=\beta r^2\,\hat{\mathbf r}\), \(\beta=1.5\times10^{-4}\ \text{C·m}^{-4}\), \(R=0.050\ \text{m}\). Using bound charge and Gauss's law, find \(\mathbf{E}(r)\) inside (\(r<R\)); there is no free charge.
Solution
\(\rho_b=-\frac{1}{r^2}\frac{d}{dr}(r^2\cdot\beta r^2)=-\frac{1}{r^2}\frac{d}{dr}(\beta r^4)=-\frac{1}{r^2}(4\beta r^3)=-4\beta r\). Enclosed bound charge in radius \(r\): \(Q_b(r)=\int_0^r(-4\beta r')4\pi r'^2\,dr'=-16\pi\beta\int_0^r r'^3 dr'=-4\pi\beta r^4\). Gauss for \(\mathbf E\): \(E\cdot4\pi r^2=Q_b/\varepsilon_0\Rightarrow E=\frac{-4\pi\beta r^4}{4\pi\varepsilon_0 r^2}=-\frac{\beta r^2}{\varepsilon_0}\). Direction \(\hat{\mathbf r}\), so \(\mathbf E=-\frac{\beta r^2}{\varepsilon_0}\hat{\mathbf r}\). This equals \(-\mathbf P/\varepsilon_0\), confirming \(\mathbf D=0\) (no free charge, spherical symmetry). At \(r=R\): \(E=-\frac{(1.5\times10^{-4})(0.050)^2}{8.85\times10^{-12}}=-4.24\times10^{4}\ \text{V·m}^{-1}\).
(C) Show that for any finite polarized body the total bound charge vanishes, \(\int_{\mathcal V}\rho_b\,d^3r+\oint_{\partial\mathcal V}\sigma_b\,da=0\), and interpret physically.
Solution
\(\int_{\mathcal V}\rho_b\,d^3r=\int_{\mathcal V}(-\nabla\cdot\mathbf P)\,d^3r=-\oint_{\partial\mathcal V}\mathbf P\cdot\hat{\mathbf n}\,da\) by the divergence theorem. The surface term is \(\oint_{\partial\mathcal V}\sigma_b\,da=\oint_{\partial\mathcal V}\mathbf P\cdot\hat{\mathbf n}\,da\). Their sum is \(-\oint\mathbf P\cdot\hat{\mathbf n}\,da+\oint\mathbf P\cdot\hat{\mathbf n}\,da=0\). Physically: polarization only redistributes existing atomic charge into aligned dipoles; it neither creates nor destroys net charge, so a body that was neutral before polarizing stays neutral. Every bound volume charge is exactly compensated by bound surface charge (and vice versa).