Dielectrics, Bound Charge and the D Field

Derivation

Dielectrics, Bound Charge and the D Field

D-048 Home PU-102 Threads fields · matter Depends on Multipole Expansion of the Potential, Differential Gauss's Law via the Divergence Theorem

Statement

A dielectric with polarization field \(\vec{P}(\vec{r})\) (electric dipole moment per unit volume) is electrostatically equivalent to a bound volume charge density \(\rho_b=-\nabla\cdot\vec{P}\) together with a bound surface charge density \(\sigma_b=\vec{P}\cdot\hat{n}\). Consequently the auxiliary field \(\vec{D}\equiv\varepsilon_0\vec{E}+\vec{P}\) obeys Gauss's law sourced by the free charge alone, \(\nabla\cdot\vec{D}=\rho_f\), with integral form \(\oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}}\).

Why it matters

Real matter is not empty space with charges pasted in: an applied field displaces bound electrons and stretches molecules, and the resulting polarization itself contributes to the field. Splitting the total charge into "free" charge we control (on electrodes, in doping) and "bound" charge the material supplies in response lets us solve for fields inside matter without tracking \(10^{23}\) atoms.

The field \(\vec{D}\) is engineered precisely so that its divergence forgets the bound charge. This is what makes boundary-value problems in capacitors, coaxial cables and dielectric interfaces tractable, and it underlies the constitutive description of every insulator, from mica to cell membranes.

Assumptions

The medium is describable by a polarization density \(\vec{P}\) = dipole moment per unit volume.If the charge distribution has no well-defined local dipole density (free charges, or scales below a molecule where "per unit volume" is meaningless) the multipole picture that produces \(\rho_b=-\nabla\cdot\vec{P}\) does not hold.

Electrostatics: charges are static, \(\partial\vec{P}/\partial t=0\).A time-varying polarization carries a polarization current \(\vec{J}_p=\partial\vec{P}/\partial t\); charge conservation then couples bound charge to that current and the static bookkeeping alone is incomplete.

The potential of the dielectric is a superposition of point-dipole potentials, truncated at the dipole term.Retaining quadrupole and higher moments adds gradient-of-gradient source terms (\(\sim\nabla\nabla\!:\!\mathbf{Q}\)); dropping this assumption means \(\rho_b=-\nabla\cdot\vec{P}\) is only the leading contribution to the bound charge.

Fields and \(\vec{P}\) are macroscopic (spatially averaged) quantities, not microscopic atomic fields.Without coarse-graining over many molecules, \(\vec{E}\) and \(\vec{P}\) fluctuate wildly between and inside atoms, and neither the divergence relation nor a smooth \(\vec{D}\) is meaningful.

Derivation

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

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

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

Gradient identity taken with respect to the source coordinate \(\vec{r}\,'\); the primed gradient of \(1/\left|\vec{r}-\vec{r}\,'\right|\) points toward the field point (the sign is opposite to the unprimed gradient). A

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

Substitute step 2 into step 1. A

\[ \nabla'\!\cdot\!\left(\frac{\vec{P}}{\left|\vec{r}-\vec{r}\,'\right|}\right)=\vec{P}\cdot\nabla'\!\left(\frac{1}{\left|\vec{r}-\vec{r}\,'\right|}\right)+\frac{1}{\left|\vec{r}-\vec{r}\,'\right|}\left(\nabla'\!\cdot\vec{P}\right) \]

Product rule for the divergence of (scalar \(\times\) vector), used to convert the integrand into a total divergence plus a remainder. B

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

Solve step 4 for \(\vec{P}\cdot\nabla'(\cdots)\) and insert into step 3. B

\[ V=\frac{1}{4\pi\varepsilon_0}\left[\oint_{\mathcal{S}'}\frac{\vec{P}\cdot\hat{n}}{\left|\vec{r}-\vec{r}\,'\right|}\,da'+\int_{\mathcal{V}'}\frac{-\nabla'\!\cdot\vec{P}}{\left|\vec{r}-\vec{r}\,'\right|}\,d^3r'\right] \]

Divergence theorem turns the first volume integral into a surface integral over the boundary \(\mathcal{S}'\) with outward normal \(\hat{n}\). B

\[ \sigma_b\equiv\vec{P}\cdot\hat{n},\qquad \rho_b\equiv-\nabla\cdot\vec{P} \]

Match each kernel \(1/\left|\vec{r}-\vec{r}\,'\right|\) against the Coulomb potential of a surface/volume charge: whatever multiplies it is that charge density. A

\[ V(\vec{r})=\frac{1}{4\pi\varepsilon_0}\left[\oint_{\mathcal{S}'}\frac{\sigma_b}{\left|\vec{r}-\vec{r}\,'\right|}\,da'+\int_{\mathcal{V}'}\frac{\rho_b}{\left|\vec{r}-\vec{r}\,'\right|}\,d^3r'\right] \]

The polarized object is exactly a set of real charges \(\rho_b,\sigma_b\): bound charge is genuine charge, not a mathematical fiction. A

\[ \nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_0}=\frac{\rho_f+\rho_b}{\varepsilon_0} \]

Differential Gauss's law is sourced by all charge; split the total density into free plus bound (from the Gauss-divergence result). A

\[ \varepsilon_0\left(\nabla\cdot\vec{E}\right)=\rho_f-\nabla\cdot\vec{P} \]

Multiply by \(\varepsilon_0\) and substitute \(\rho_b=-\nabla\cdot\vec{P}\) from step 7. A

\[ \nabla\cdot\left(\varepsilon_0\vec{E}+\vec{P}\right)=\rho_f \]

Bring \(\nabla\cdot\vec{P}\) to the left; linearity of the divergence lets the two terms combine under one \(\nabla\cdot\). A

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

Define the displacement field as the quantity whose divergence is the free charge; the definition is what removes bound charge from view. A

\[ \int_{\mathcal{V}}\nabla\cdot\vec{D}\,d^3r=\oint_{\mathcal{S}}\vec{D}\cdot d\vec{a}=\int_{\mathcal{V}}\rho_f\,d^3r=Q_{f,\text{enc}} \]

Integrate step 12 over an arbitrary volume and apply the divergence theorem. Valid even across surfaces where \(\vec{P}\) jumps, since only \(\nabla\cdot\vec{D}=\rho_f\) in the distributional sense was used. C

Result

\[ \rho_b=-\nabla\cdot\vec{P},\quad \sigma_b=\vec{P}\cdot\hat{n},\quad \vec{D}=\varepsilon_0\vec{E}+\vec{P},\quad \nabla\cdot\vec{D}=\rho_f,\quad \oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}} \]

Reading. Where the polarization "piles up" (nonzero divergence), positive charge has been pushed out faster than it arrived, leaving a net negative bound charge — hence the minus sign. At a surface, \(\vec{P}\cdot\hat{n}\) counts the dipole heads left stranded on the boundary. The displacement \(\vec{D}\) is the specific combination of the true field \(\vec{E}\) and the material response \(\vec{P}\) whose flux through any closed surface counts only the free charge you placed — the bound charge cancels internally.

Units check. \([\vec{P}]=\)C·m/m³ = C·m⁻² (dipole moment per volume). Then \([\nabla\cdot\vec{P}]=\)C·m⁻³ = charge density, and \([\vec{P}\cdot\hat{n}]=\)C·m⁻² = surface charge density. \([\varepsilon_0\vec{E}]=\)(C²·N⁻¹·m⁻²)(N·C⁻¹) = C·m⁻², matching \([\vec{P}]\), so \([\vec{D}]=\)C·m⁻² and \([\nabla\cdot\vec{D}]=\)C·m⁻³ \(=[\rho_f]\).

Limiting cases

Uniform polarization (\(\vec{P}=\text{const}\)): \(\nabla\cdot\vec{P}=0\), so \(\rho_b=0\) and all bound charge sits on the surface as \(\sigma_b=\vec{P}\cdot\hat{n}\) — e.g. a uniformly polarized slab or sphere.
Vacuum / no matter (\(\vec{P}=0\)): \(\vec{D}=\varepsilon_0\vec{E}\) and \(\nabla\cdot\vec{D}=\rho_f\) collapses to ordinary Gauss's law.
Linear isotropic dielectric (\(\vec{P}=\varepsilon_0\chi_e\vec{E}\)): \(\vec{D}=\varepsilon_0(1+\chi_e)\vec{E}=\varepsilon\vec{E}\); inside a region containing only free charge, \(\rho_b=-\dfrac{\chi_e}{1+\chi_e}\,\rho_f\).
Conductor limit (\(\chi_e\to\infty,\ \varepsilon\to\infty\)): \(\vec{E}\to0\) inside, bound surface charge screens the free charge completely.

Breaks when

Time-dependent polarization. With \(\partial\vec{P}/\partial t\neq0\) a polarization current \(\vec{J}_p=\partial\vec{P}/\partial t\) flows; charge conservation still holds (\(\nabla\cdot\vec{J}_p=-\partial\rho_b/\partial t\)), but the electrostatic potential of step 1 is no longer valid and one must use the full Maxwell equations. \(\nabla\cdot\vec{D}=\rho_f\) survives, but \(\nabla\times\vec{E}\neq0\).
Strongly inhomogeneous fields (nonlocal / gradient response). When \(\vec{P}\) at a point depends on \(\vec{E}\) elsewhere (atomic-scale gradients, near interfaces, Thomas–Fermi screening in metals) the simple local dipole density fails, and higher multipoles (a quadrupole density) contribute bound charge beyond \(-\nabla\cdot\vec{P}\).
Nonlinear / hysteretic media (ferroelectrics). If \(\vec{P}\) is not single-valued in \(\vec{E}\), then \(\vec{D}=\varepsilon\vec{E}\) is meaningless; \(\nabla\cdot\vec{D}=\rho_f\) still holds as bookkeeping but no longer closes the boundary-value problem without the material's \(\vec{P}(\vec{E})\) history.
Below the coarse-graining scale. At sub-molecular resolution there is no smooth \(\vec{P}\); only discrete microscopic charges exist, and the macroscopic \(\vec{D}\) loses meaning.

Failure modes

Sign flip on \(\rho_b\). Writing \(\rho_b=+\nabla\cdot\vec{P}\). The minus is not cosmetic — it comes from the product rule (step 5), and getting it wrong reverses every bound-charge field.
Thinking \(\vec{D}\) has no curl. \(\nabla\times\vec{D}=\nabla\times\vec{P}\neq0\) in general, so \(\vec{D}\) is not derivable from a potential, and Gauss's law for \(\vec{D}\) alone does not determine it without a constitutive relation and symmetry.
Using \(\vec{D}=\varepsilon\vec{E}\) in a nonlinear or anisotropic medium. The scalar \(\varepsilon\) exists only for linear isotropic materials; in crystals \(\varepsilon\) is a tensor and \(\vec{D}\) need not be parallel to \(\vec{E}\).
Forgetting \(\sigma_b\) at interfaces. Applying \(\rho_b=-\nabla\cdot\vec{P}\) in the bulk while dropping the surface term loses charge; \(\oint\sigma_b\,da+\int\rho_b\,d^3r=0\) must hold.
Treating bound charge as fictitious. \(\rho_b\) produces real \(\vec{E}\) fields and real forces; it is simply charge you did not place by hand.
Confusing \(\varepsilon_r\) with \(\chi_e\). \(\varepsilon_r=1+\chi_e\), not \(\chi_e\).

Discussion

The deepest content of this derivation is the identification in steps 7–8: the potential of a polarized body is algebraically identical to that of a body carrying charge densities \(-\nabla\cdot\vec{P}\) and \(\vec{P}\cdot\hat{n}\). Nothing is approximated in that identification, given the dipole truncation. The bound charge is as real as any other — it is the accumulated displacement of atomic charge — and the free/bound split is a choice of bookkeeping, not a division of physical reality: what we control (free charge on electrodes) versus what the material supplies in response.

The field \(\vec{D}\) is then a construction of convenience: we bundle the awkward \(-\nabla\cdot\vec{P}\) source back into the field so its divergence sees only \(\rho_f\). This is useful because \(\rho_f\) is what an experimenter specifies. But it comes at a price: \(\vec{D}\) is not a fundamental field. Its curl is set by \(\nabla\times\vec{P}\), so unlike \(\vec{E}\) it is generally not curl-free and cannot be obtained from Gauss's law plus symmetry alone except in configurations where \(\nabla\times\vec{P}=0\).

The boundary conditions that fall out — \(D_{\perp,\text{above}}-D_{\perp,\text{below}}=\sigma_f\) and \(E_{\parallel,\text{above}}=E_{\parallel,\text{below}}\) — are the workhorses of dielectric problems: the normal component of \(\vec{D}\) is continuous across an uncharged interface while the tangential \(\vec{E}\) is always continuous. With a constitutive relation they close every capacitor and interface problem, and their combination bends field lines at a dielectric boundary (the electrostatic analogue of refraction).

A subtlety worth stating precisely: \(\rho_b=-\nabla\cdot\vec{P}\) is exact only if \(\vec{P}\) captures the full leading dipole density and higher moments are negligible. In a rigorous microscopic-to-macroscopic averaging (Russakoff), the macroscopic charge density is \(\langle\rho_{\text{micro}}\rangle=\rho_f-\nabla\cdot\vec{P}+\partial_i\partial_j Q_{ij}-\cdots\), where \(Q_{ij}\) is a quadrupole density. The auxiliary field is more properly \(D_i=\varepsilon_0 E_i+P_i-\partial_j Q_{ij}+\cdots\); the familiar \(\vec{D}=\varepsilon_0\vec{E}+\vec{P}\) is the electric-dipole approximation, excellent for ordinary dielectrics but not a fundamental law.

Common misconceptions. \(\vec{D}\) is not "the field in a dielectric" — that is \(\vec{E}\), and the force on a test charge is \(q\vec{E}\), never \(q\vec{D}/\varepsilon_0\). Gauss's law for \(\vec{D}\) does not make \(\vec{D}\) depend only on free charge in general geometry (it depends on \(\vec{E}\) and \(\vec{P}\), hence on bound charge too); only its flux is blind to bound charge. And "no free charge" does not mean "no \(\vec{D}\)": a uniformly polarized sphere in vacuum has \(\rho_f=0\) everywhere yet nonzero \(\vec{D}\) inside.

Worked examples

Uniformly polarized dielectric sphere.

Sphere radius \(R=2.0\ \text{cm}\), uniform \(\vec{P}=P\hat{z}\) with \(P=3.0\times10^{-6}\ \text{C·m}^{-2}\). Find \(\rho_b\), \(\sigma_b(\theta)\), and the total bound charge on the upper hemisphere. A

\[ \rho_b=-\nabla\cdot\vec{P}=0\quad(\vec{P}\ \text{constant}) \]

A uniform vector field has zero divergence. A

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

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

\[ Q_{\text{upper}}=\int_0^{2\pi}\!\!\int_0^{\pi/2}P\cos\theta\;R^2\sin\theta\,d\theta\,d\phi=2\pi PR^2\!\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\pi PR^2 \]

Integrate \(\sigma_b\) over the upper hemisphere; \(\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\tfrac12\). B

\[ Q_{\text{upper}}=\pi(3.0\times10^{-6})(0.020)^2=\pi(3.0\times10^{-6})(4.0\times10^{-4})=3.8\times10^{-9}\ \text{C} \]

Substitute numbers; units C·m⁻² · m² = C. A

\[ \rho_b=0,\quad \sigma_b=P\cos\theta=(3.0\ \mu\text{C·m}^{-2})\cos\theta,\quad Q_{\text{upper}}\approx+3.8\ \text{nC} \]

Reading. All bound charge is on the surface, positive on the top cap and negative on the bottom, with equal magnitude so the sphere stays neutral. Units check: C·m⁻²·m² = C.

Point free charge in a linear dielectric.

Point free charge \(Q_f=5.0\ \text{nC}\) embedded in an infinite linear dielectric with \(\varepsilon_r=1+\chi_e=4.0\). Find \(\vec{D}(r)\), \(\vec{E}(r)\), and the total bound charge \(q_b\) hugging the point charge. A

\[ \oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}}\;\Rightarrow\;D(4\pi r^2)=Q_f\;\Rightarrow\;\vec{D}=\frac{Q_f}{4\pi r^2}\hat{r} \]

Spherical symmetry: \(\vec{D}\) is radial and constant on the Gaussian sphere; only free charge sources it. A

\[ \vec{E}=\frac{\vec{D}}{\varepsilon}=\frac{\vec{D}}{\varepsilon_0\varepsilon_r}=\frac{Q_f}{4\pi\varepsilon_0\varepsilon_r r^2}\hat{r} \]

Linear isotropic constitutive relation \(\vec{D}=\varepsilon\vec{E}\). The field is reduced by \(1/\varepsilon_r\) versus vacuum. A

\[ q_b=-\frac{\chi_e}{\varepsilon_r}Q_f=-\frac{\varepsilon_r-1}{\varepsilon_r}Q_f \]

The point bound charge is \(-\nabla\cdot\vec{P}\) integrated over an infinitesimal shell; with \(\vec{P}=\varepsilon_0\chi_e\vec{E}\) this is the standard screening result. B

\[ q_b=-\frac{3.0}{4.0}(5.0\ \text{nC})=-3.75\ \text{nC},\qquad E(1\ \text{cm})=\frac{5.0\times10^{-9}}{4\pi(8.85\times10^{-12})(4.0)(0.01)^2}\approx1.1\times10^{4}\ \text{N·C}^{-1} \]

Insert numbers; \(\varepsilon_0=8.85\times10^{-12}\ \text{F·m}^{-1}\). A

\[ \vec{D}=\frac{Q_f}{4\pi r^2}\hat{r},\quad \vec{E}=\frac{\vec{D}}{\varepsilon_0\varepsilon_r},\quad q_b=-3.75\ \text{nC} \]

Reading. The dielectric screens the point charge: a shell of \(-3.75\) nC bound charge surrounds it, so the effective charge seen far away is \(Q_f/\varepsilon_r=1.25\) nC, exactly the factor-4 reduction of \(\vec{E}\). Units check: \([\vec{D}]=\)C·m⁻², \([\vec{E}]=\)C·m⁻²/(C²·N⁻¹·m⁻²) = N·C⁻¹.

Problems

(A) A slab fills \(0\le z\le d\) with polarization \(\vec{P}=P_0(z/d)\hat{z}\). Find \(\rho_b\) and the two surface densities, and verify neutrality.
Solution
\(\rho_b=-\nabla\cdot\vec{P}=-\partial_z(P_0 z/d)=-P_0/d\) (uniform, negative). At \(z=d\) (top, \(\hat{n}=+\hat{z}\)): \(\sigma_b=\vec{P}\cdot\hat{n}=P_0\). At \(z=0\) (bottom, \(\hat{n}=-\hat{z}\)): \(\vec{P}(0)=0\) so \(\sigma_b=0\). Total per unit area: \(-(P_0/d)\,d+P_0=0\), neutral.
(A) A dielectric sphere (radius \(R\), permittivity \(\varepsilon\)) contains uniform free charge density \(\rho_f\). Find \(\vec{D}\) and \(\vec{E}\) inside.
Solution
By symmetry \(\oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}}\): \(D(4\pi r^2)=\rho_f\tfrac43\pi r^3\Rightarrow \vec{D}=\tfrac13\rho_f r\,\hat{r}\) for \(r\le R\). Then \(\vec{E}=\vec{D}/\varepsilon=\dfrac{\rho_f r}{3\varepsilon}\hat{r}\). Note \(\vec{D}\) is independent of \(\varepsilon\) because only free charge and geometry enter Gauss's law for \(\vec{D}\).
(B) Show explicitly that the total bound charge of any finite polarized body is zero: \(\int_{\mathcal V}\rho_b\,d^3r+\oint_{\mathcal S}\sigma_b\,da=0\).
Solution
\(\int_{\mathcal V}\rho_b\,d^3r=-\int_{\mathcal V}\nabla\cdot\vec{P}\,d^3r=-\oint_{\mathcal S}\vec{P}\cdot d\vec{a}\) (divergence theorem). And \(\oint_{\mathcal S}\sigma_b\,da=\oint_{\mathcal S}\vec{P}\cdot\hat{n}\,da=+\oint_{\mathcal S}\vec{P}\cdot d\vec{a}\). The two are equal and opposite, so the sum vanishes. Physically, polarization merely rearranges internal charge and creates no net charge.
(B) A parallel-plate capacitor, plate area \(A=100\ \text{cm}^2\), gap \(d=1.0\ \text{mm}\), holds free surface charge \(\sigma_f=2.0\times10^{-6}\ \text{C·m}^{-2}\) and is filled with a dielectric \(\varepsilon_r=5.0\). Find \(D\), \(E\), \(P\), \(\sigma_b\) and the voltage.
Solution
\(D=\sigma_f=2.0\times10^{-6}\ \text{C·m}^{-2}\) (pillbox: \(\oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}}\)). \(E=D/(\varepsilon_0\varepsilon_r)=2.0\times10^{-6}/(8.85\times10^{-12}\cdot5.0)=4.52\times10^{4}\ \text{V·m}^{-1}\). \(P=D-\varepsilon_0E=2.0\times10^{-6}-8.85\times10^{-12}\cdot4.52\times10^{4}=1.60\times10^{-6}\ \text{C·m}^{-2}\). \(\sigma_b=\vec{P}\cdot\hat{n}=1.60\times10^{-6}\ \text{C·m}^{-2}\) (sign opposite the adjacent free charge). \(V=Ed=4.52\times10^{4}\cdot1.0\times10^{-3}=45.2\ \text{V}\). Check: \(\sigma_f-\sigma_b=0.40\times10^{-6}=\sigma_f/\varepsilon_r\).
(C) A long cylinder (radius \(a\)) carries frozen-in polarization \(\vec{P}=ks\,\hat{s}\) (radial, \(s\) the cylindrical radius, \(k\) constant), no free charge. Find \(\vec{E}\) inside using bound charge, then verify with \(\vec{D}\).
Solution
\(\rho_b=-\nabla\cdot\vec{P}=-\dfrac{1}{s}\partial_s(s\cdot ks)=-\dfrac{1}{s}(2ks)=-2k\). Surface \(\sigma_b=\vec{P}(a)\cdot\hat{s}=ka\). Gauss with a coaxial cylinder of radius \(s<a\), length \(L\): \(E(2\pi sL)=\rho_b(\pi s^2 L)/\varepsilon_0\Rightarrow E=\rho_b s/(2\varepsilon_0)=-ks/\varepsilon_0\) (radial, inward). Verify with \(\vec{D}\): no free charge plus symmetry give \(\vec{D}=0\) everywhere; then \(\vec{E}=(\vec{D}-\vec{P})/\varepsilon_0=-\vec{P}/\varepsilon_0=-\dfrac{ks}{\varepsilon_0}\hat{s}\). This warns that \(\vec{D}=0\) does not imply \(\vec{E}=0\).