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Physics · Electromagnetism and Electrodynamics

Dielectrics, Polarisation and the Electric Displacement Field

A polarized dielectric with polarization P⃗\vec{P} (dipole moment per unit volume) is equivalent to bound charges ρb=−∇⋅P⃗\rho_b=-\nabla\cdot\vec{P} and σb=P⃗⋅n^\sigma_b=\vec{P}\cdot\hat{n}; the displacement D⃗=ε0E⃗+P⃗\vec{D}=\varepsilon_0\vec{E}+\vec{P} obeys ∇⋅D⃗=ρf\nabla\cdot\vec{D}=\rho_f but is not curl-free (∇×D⃗=∇×P⃗\nabla\times\vec{D}=\nabla\times\vec{P}). For linear media D⃗=εE⃗\vec{D}=\varepsilon\vec{E} with ε=ε0(1+χe)\varepsilon=\varepsilon_0(1+\chi_e); D⊥D_\perp jumps by σf\sigma_f and E⃗∥\vec{E}_\parallel is continuous; a dielectric sphere in E0z^E_0\hat{z} has uniform interior field 3ε0ε+2ε0E0\tfrac{3\varepsilon_0}{\varepsilon+2\varepsilon_0}E_0; the energy is 12∫D⃗⋅E⃗ dτ\tfrac12\int\vec{D}\cdot\vec{E}\,d\tau and dielectrics are drawn into strong-field regions.

Dielectrics: Polarization, Bound Charge and the Displacement Field

So far the sources of the electrostatic field have been "free" charges — charges we place and control. Ordinary matter contains a vastly larger number of charges bound into atoms and molecules. In an external field these bound charges shift slightly, each atom acquiring a small dipole moment, and the material becomes polarized. This chapter develops the electrostatics of polarizable matter in the manner of Griffiths' Chapter 4: the polarization field P⃗\vec{P}, the bound charges it is equivalent to, the displacement D⃗\vec{D} that lets Gauss's law be written in terms of free charge alone, the constitutive relation for linear dielectrics, and the modified boundary conditions and energy expression.

The central bookkeeping idea is this. A polarized object is, as far as its external and internal fields are concerned, exactly equivalent to a volume charge density ρb=−∇⋅P⃗\rho_b = -\nabla\cdot\vec{P} plus a surface charge density σb=P⃗⋅n^\sigma_b = \vec{P}\cdot\hat{n}. Once you accept that, everything reduces to electrostatics you already know, with ρb\rho_b added to ρf\rho_f. The vector D⃗=ε0E⃗+P⃗\vec{D} = \varepsilon_0\vec{E} + \vec{P} is then defined precisely so that its divergence sees only the free charge: ∇⋅D⃗=ρf\nabla\cdot\vec{D} = \rho_f. It is a convenience, not a new physical field — and it can mislead, because D⃗\vec{D} is not curl-free in general.

How to read this chapter. Hold on to three relations: ρb=−∇⋅P⃗\rho_b = -\nabla\cdot\vec{P} and σb=P⃗⋅n^\sigma_b = \vec{P}\cdot\hat{n} (the bound charges), ∇⋅D⃗=ρf\nabla\cdot\vec{D} = \rho_f (Gauss for D⃗\vec{D}), and D⃗=εE⃗\vec{D} = \varepsilon\vec{E} with ε=ε0(1+χe)\varepsilon = \varepsilon_0(1+\chi_e) (linear media). The recurring trap is treating D⃗\vec{D} as if it were an electric field: it obeys Gauss's law with free charge only — so D⃗\vec{D}-field lines can begin and end only on free charge (or close on themselves) — but its curl is ∇×P⃗≠0\nabla\times\vec{P} \ne 0 in general, so unlike E⃗\vec{E} it has no scalar potential and ∮D⃗⋅dℓ⃗≠0\oint\vec{D}\cdot d\vec{\ell}\ne 0 in general.

1. Polarization

When a dielectric is placed in a field, two microscopic things can happen: non-polar atoms/molecules develop an induced dipole moment p⃗=αE⃗\vec{p} = \alpha\vec{E} (electron cloud displaced against the nucleus), and polar molecules (which already have a permanent p⃗\vec{p}) partially align with the field against thermal disorder. Either way, the macroscopic result is a dipole moment per unit volume,

P⃗(r⃗)=dp⃗dτ=n⟨p⃗⟩\vec{P}(\vec{r}) = \frac{d\vec{p}}{d\tau} = n\langle\vec{p}\rangle

with nn the number density of dipoles. P⃗\vec{P} is the fundamental field of this chapter; everything else is derived from it.

2. The Field of a Polarized Object: Bound Charges

The potential of a single dipole p⃗\vec{p} at the origin is V=14πε0p⃗⋅r^r2V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}. Summing over all the dipoles in a polarized body, with p⃗→P⃗ dτ′\vec{p} \to \vec{P}\,d\tau',

V(r⃗)=14πε0∫VP⃗(r⃗ ′)⋅η^η2 dτ′V(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal{V}} \frac{\vec{P}(\vec{r}\,')\cdot\hat{\eta}}{\eta^2}\,d\tau'

Using η^/η2=∇′(1/η)\hat{\eta}/\eta^2 = \nabla'(1/\eta) and integrating by parts (product rule for the divergence, then the divergence theorem),

V(r⃗)=14πε0∮SP⃗⋅n^η da′+14πε0∫V−∇′⋅P⃗η dτ′V(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\oint_{\mathcal{S}} \frac{\vec{P}\cdot\hat{n}}{\eta}\,da' + \frac{1}{4\pi\varepsilon_0}\int_{\mathcal{V}} \frac{-\nabla'\cdot\vec{P}}{\eta}\,d\tau'

The first term is the potential of a surface charge σb=P⃗⋅n^\sigma_b = \vec{P}\cdot\hat{n}; the second is that of a volume charge ρb=−∇⋅P⃗\rho_b = -\nabla\cdot\vec{P}. So:

  ρb=−∇⋅P⃗,σb=P⃗⋅n^  \boxed{\;\rho_b = -\nabla\cdot\vec{P}, \qquad \sigma_b = \vec{P}\cdot\hat{n}\;}

These "bound charges" are entirely real accumulations of charge — the ends of the little dipoles that fail to cancel their neighbours — not a mathematical fiction. Their total is zero: ∫ρb dτ+∮σb da=−∫∇⋅P⃗ dτ+∮P⃗⋅n^ da=0\displaystyle\int\rho_b\,d\tau + \oint\sigma_b\,da = -\int\nabla\cdot\vec{P}\,d\tau + \oint\vec{P}\cdot\hat{n}\,da = 0 by the divergence theorem, as it must be for a neutral dielectric.

Worked example — uniformly polarized sphere. A sphere of radius RR has uniform P⃗=Pz^\vec{P} = P\hat{z}. Then ρb=−∇⋅P⃗=0\rho_b = -\nabla\cdot\vec{P} = 0 inside, and σb=P⃗⋅n^=Pcos⁡θ\sigma_b = \vec{P}\cdot\hat{n} = P\cos\theta on the surface. That surface density is exactly the pattern that produces a uniform field inside: E⃗in=−P⃗3ε0\vec{E}_{\text{in}} = -\dfrac{\vec{P}}{3\varepsilon_0}, and a pure dipole field outside with moment p⃗=43πR3P⃗\vec{p} = \tfrac{4}{3}\pi R^3\vec{P} (the total dipole moment of the sphere). This −P⃗/3ε0-\vec{P}/3\varepsilon_0 interior field is the origin of the "depolarizing field" and the factor of 3 that recurs throughout dielectric problems.

3. The Displacement Field

Inside a dielectric the total charge is ρ=ρf+ρb\rho = \rho_f + \rho_b, so Gauss's law reads ∇⋅E⃗=(ρf+ρb)/ε0=(ρf−∇⋅P⃗)/ε0\nabla\cdot\vec{E} = (\rho_f + \rho_b)/\varepsilon_0 = (\rho_f - \nabla\cdot\vec{P})/\varepsilon_0. Rearranging,

∇⋅(ε0E⃗+P⃗)=ρf\nabla\cdot(\varepsilon_0\vec{E} + \vec{P}) = \rho_f

Define the electric displacement

  D⃗≡ε0E⃗+P⃗,∇⋅D⃗=ρf,∯D⃗⋅da⃗=Qf,enc  \boxed{\;\vec{D} \equiv \varepsilon_0\vec{E} + \vec{P}, \qquad \nabla\cdot\vec{D} = \rho_f, \qquad \oiint\vec{D}\cdot d\vec{a} = Q_{f,\text{enc}}\;}

The utility is real: in a problem with enough symmetry, you can find D⃗\vec{D} from the free charge alone, without knowing the polarization, then get E⃗\vec{E} from a constitutive relation. But D⃗\vec{D} is not the whole story:

∇×D⃗=ε0(∇×E⃗)+∇×P⃗=∇×P⃗\nabla\times\vec{D} = \varepsilon_0(\nabla\times\vec{E}) + \nabla\times\vec{P} = \nabla\times\vec{P}

which is generally non-zero. There is no "potential for D⃗\vec{D}" and ∮D⃗⋅dℓ⃗≠0\oint\vec{D}\cdot d\vec{\ell}\ne 0 in general — but its sources are still only free charge: D⃗\vec{D}-field lines can start and stop only on free charge (or close on themselves), since ∇⋅D⃗=ρf\nabla\cdot\vec{D} = \rho_f exactly. Treat D⃗\vec{D} strictly as a bookkeeping device for Gauss's law.

4. Linear Dielectrics: Susceptibility and Permittivity

For many materials, in fields that are not too strong, the polarization is proportional to the field that produces it:

P⃗=ε0χe E⃗\vec{P} = \varepsilon_0\chi_e\,\vec{E}

with χe\chi_e the (dimensionless) electric susceptibility. Note E⃗\vec{E} here is the total macroscopic field, including the contribution of P⃗\vec{P} itself. Substituting into the definition of D⃗\vec{D}:

D⃗=ε0(1+χe)E⃗≡εE⃗,ε=ε0(1+χe),εr≡εε0=1+χe\vec{D} = \varepsilon_0(1+\chi_e)\vec{E} \equiv \varepsilon\vec{E}, \qquad \varepsilon = \varepsilon_0(1+\chi_e), \qquad \varepsilon_r \equiv \frac{\varepsilon}{\varepsilon_0} = 1+\chi_e

ε\varepsilon is the permittivity and εr\varepsilon_r the relative permittivity or dielectric constant. In a linear homogeneous dielectric with free charge ρf\rho_f, ∇⋅E⃗=ρf/ε\nabla\cdot\vec{E} = \rho_f/\varepsilon: the vacuum results carry over with ε0→ε\varepsilon_0 \to \varepsilon. Also ρb=−∇⋅P⃗=−χe1+χeρf\rho_b = -\nabla\cdot\vec{P} = -\dfrac{\chi_e}{1+\chi_e}\rho_f, so bound volume charge appears only where there is free volume charge; a homogeneous linear dielectric with ρf=0\rho_f = 0 has all its bound charge on surfaces.

Worked example — capacitor with a dielectric. Fill a parallel-plate capacitor (free surface charge ±σf\pm\sigma_f) with a linear dielectric ε\varepsilon. A Gaussian pillbox gives D=σfD = \sigma_f (only free charge counts), so E=σf/εE = \sigma_f/\varepsilon, smaller by εr\varepsilon_r than the vacuum value. The voltage drops by εr\varepsilon_r, so the capacitance rises: C=εrCvacC = \varepsilon_r C_{\text{vac}}. The bound surface charge on the dielectric face is σb=P=ε0χeE=σf χe/εr\sigma_b = P = \varepsilon_0\chi_e E = \sigma_f\,\chi_e/\varepsilon_r, opposing the free charge and thereby weakening the field.

5. Boundary Conditions with Dielectrics

At an interface between two media, integrate the two field equations across a pillbox and a thin loop:

D⊥above−D⊥below=σf,E⃗∥above−E⃗∥below=0⃗D_\perp^{\text{above}} - D_\perp^{\text{below}} = \sigma_f, \qquad \vec{E}_\parallel^{\text{above}} - \vec{E}_\parallel^{\text{below}} = \vec{0}

The normal component of D⃗\vec{D} jumps by the free surface charge; the tangential component of E⃗\vec{E} is continuous. Equivalently, D⊥D_\perp is continuous across an interface with no free charge, and E⊥E_\perp jumps (by σb/ε0\sigma_b/\varepsilon_0). For linear media these combine into the refraction rule ε1E1⊥=ε2E2⊥\varepsilon_1 E_{1\perp} = \varepsilon_2 E_{2\perp} (no free charge), which bends field lines at a dielectric interface just as Snell's law bends light rays.

6. Boundary Value Problems in Dielectrics

The strategy: solve Laplace's equation for VV in each linear dielectric region (∇2V=0\nabla^2 V = 0 where ρf=0\rho_f = 0), then match at interfaces using continuity of VV and continuity of D⊥=−ε ∂V/∂nD_\perp = -\varepsilon\,\partial V/\partial n.

Worked example — dielectric sphere in a uniform field. A sphere of radius RR, permittivity ε\varepsilon, sits in a uniform field E0z^E_0\hat{z} in vacuum. Inside, Vin=−Arcos⁡θV_{\text{in}} = -A r\cos\theta (regular at the origin, ℓ=1\ell=1 only). Outside, Vout=−E0rcos⁡θ+Br2cos⁡θV_{\text{out}} = -E_0 r\cos\theta + \dfrac{B}{r^2}\cos\theta. Matching VV and ε ∂V/∂r\varepsilon\,\partial V/\partial r at r=Rr = R gives

E⃗in=3ε0ε+2ε0 E0 z^,B=ε−ε0ε+2ε0 R3E0\vec{E}_{\text{in}} = \frac{3\varepsilon_0}{\varepsilon + 2\varepsilon_0}\,E_0\,\hat{z}, \qquad B = \frac{\varepsilon - \varepsilon_0}{\varepsilon + 2\varepsilon_0}\,R^3 E_0

The interior field is uniform, weaker than E0E_0, reduced by the factor 3ε0/(ε+2ε0)3\varepsilon_0/(\varepsilon + 2\varepsilon_0). As ε→∞\varepsilon \to \infty this recovers the conductor result (E⃗in→0\vec{E}_{\text{in}} \to 0, B→R3E0B \to R^3 E_0). The induced dipole moment is p=4πε0B=4πε0R3E0 ε−ε0ε+2ε0p = 4\pi\varepsilon_0 B = 4\pi\varepsilon_0 R^3 E_0\,\dfrac{\varepsilon-\varepsilon_0}{\varepsilon+2\varepsilon_0} — the Clausius–Mossotti form.

7. Energy in Dielectric Systems

The work to assemble a configuration of free charges in the presence of linear dielectrics is

W=12∫ρf V dτ=12∫D⃗⋅E⃗ dτ=12∫εE2 dτW = \frac{1}{2}\int \rho_f\,V\,d\tau = \frac{1}{2}\int \vec{D}\cdot\vec{E}\,d\tau = \frac{1}{2}\int \varepsilon E^2\,d\tau

The energy density is u=12D⃗⋅E⃗u = \tfrac12\vec{D}\cdot\vec{E}. This is the energy needed to charge up the free charges; it already includes the work done polarizing the dielectric, but it does not include the energy that would be released if the dielectric molecules were pulled apart — it is the right quantity for computing forces on dielectrics at fixed free charge. Two subtleties: (i) this formula assumes linear media; for nonlinear dielectrics one must integrate ∫E⃗⋅dD⃗\int \vec{E}\cdot d\vec{D}. (ii) At fixed potential (battery connected) the force is F⃗=+∇W\vec{F} = +\nabla W, not −∇W-\nabla W, because the battery does work.

Worked example — force pulling a dielectric slab into a capacitor. A slab of dielectric constant εr\varepsilon_r is inserted a distance xx into a parallel-plate capacitor (plate width ww, total length LL, gap dd). The capacitance is C(x)=ε0wd[εrx+(L−x)]C(x) = \dfrac{\varepsilon_0 w}{d}\big[\varepsilon_r x + (L - x)\big], so dCdx=ε0w(εr−1)d>0\dfrac{dC}{dx} = \dfrac{\varepsilon_0 w(\varepsilon_r - 1)}{d} > 0. At fixed charge QQ, W=Q2/2CW = Q^2/2C and Fx=−dWdx=Q22C2dCdx>0F_x = -\dfrac{dW}{dx} = \dfrac{Q^2}{2C^2}\dfrac{dC}{dx} > 0; at fixed voltage VV, Fx=+dWdx∣V=V22dCdx=ε0w(εr−1)V22d>0F_x = +\dfrac{dW}{dx}\Big|_V = \dfrac{V^2}{2}\dfrac{dC}{dx} = \dfrac{\varepsilon_0 w(\varepsilon_r - 1)V^2}{2d} > 0. Either way the slab is pulled in. The physical origin is the fringing field acting on the bound charge at the slab's leading edge.

8. Common Pitfalls

  • Treating D⃗\vec{D} as an electric field. D⃗\vec{D} satisfies Gauss's law with free charge, but ∇×D⃗=∇×P⃗≠0\nabla\times\vec{D} = \nabla\times\vec{P} \ne 0 in general. There is no scalar potential for D⃗\vec{D} — but its field lines can still only begin and end on free charge, never on bound charge, since ∇⋅D⃗=ρf\nabla\cdot\vec{D} = \rho_f.
  • Forgetting that E⃗\vec{E} in P⃗=ε0χeE⃗\vec{P} = \varepsilon_0\chi_e\vec{E} is the total field. It includes the dielectric's own contribution — which is why the interior field of a polarized sphere is not simply E0E_0.
  • Assuming D⃗\vec{D} is determined by the free charge in general. Only high-symmetry configurations let you get D⃗\vec{D} from Qf,encQ_{f,\text{enc}} alone. In an asymmetric problem D⃗\vec{D} depends on P⃗\vec{P} too.
  • Using D⊥D_\perp continuity when there is free surface charge. D⊥D_\perp jumps by σf\sigma_f; it is continuous only where σf=0\sigma_f = 0.
  • Applying W=12∫εE2 dτW = \tfrac12\int\varepsilon E^2\,d\tau to nonlinear media. The compact energy formula relies on P⃗∝E⃗\vec{P}\propto\vec{E}; otherwise integrate ∫E⃗⋅dD⃗\int\vec{E}\cdot d\vec{D} over the charging process.
  • Sign of the force at fixed potential. With a battery attached, F⃗=+∇W∣V\vec{F} = +\nabla W|_V; using −∇W-\nabla W there gives the wrong sign because the battery supplies energy.
  • Confusing bound charge with a lack of reality. ρb\rho_b and σb\sigma_b are physical charge accumulations; you could in principle measure them by slicing the dielectric.

Summary

  • Polarization P⃗\vec{P} = dipole moment per unit volume. A polarized object is equivalent to bound charges ρb=−∇⋅P⃗\rho_b = -\nabla\cdot\vec{P} (volume) and σb=P⃗⋅n^\sigma_b = \vec{P}\cdot\hat{n} (surface), of zero total.
  • Uniformly polarized sphere: E⃗in=−P⃗/3ε0\vec{E}_{\text{in}} = -\vec{P}/3\varepsilon_0 (uniform), pure dipole field outside with p⃗=43πR3P⃗\vec{p} = \tfrac43\pi R^3\vec{P}.
  • Displacement: D⃗=ε0E⃗+P⃗\vec{D} = \varepsilon_0\vec{E} + \vec{P}, with ∇⋅D⃗=ρf\nabla\cdot\vec{D} = \rho_f and ∯D⃗⋅da⃗=Qf,enc\oiint\vec{D}\cdot d\vec{a} = Q_{f,\text{enc}}. But ∇×D⃗=∇×P⃗≠0\nabla\times\vec{D} = \nabla\times\vec{P} \ne 0 — D⃗\vec{D} is not a field with a potential.
  • Linear dielectrics: P⃗=ε0χeE⃗\vec{P} = \varepsilon_0\chi_e\vec{E}, D⃗=εE⃗\vec{D} = \varepsilon\vec{E}, ε=ε0(1+χe)\varepsilon = \varepsilon_0(1+\chi_e), εr=1+χe\varepsilon_r = 1+\chi_e. Vacuum electrostatics carries over with ε0→ε\varepsilon_0\to\varepsilon; ρb=−χe1+χeρf\rho_b = -\tfrac{\chi_e}{1+\chi_e}\rho_f.
  • Boundary conditions: D⊥above−D⊥below=σfD_\perp^{\text{above}} - D_\perp^{\text{below}} = \sigma_f, E⃗∥\vec{E}_\parallel continuous. Field lines refract: ε1E1⊥=ε2E2⊥\varepsilon_1 E_{1\perp} = \varepsilon_2 E_{2\perp} when σf=0\sigma_f = 0.
  • Dielectric sphere in E0z^E_0\hat{z}: E⃗in=3ε0ε+2ε0E0z^\vec{E}_{\text{in}} = \dfrac{3\varepsilon_0}{\varepsilon+2\varepsilon_0}E_0\hat{z} (uniform); induced p=4πε0R3E0 ε−ε0ε+2ε0p = 4\pi\varepsilon_0 R^3 E_0\,\dfrac{\varepsilon-\varepsilon_0}{\varepsilon+2\varepsilon_0}.
  • Energy: W=12∫D⃗⋅E⃗ dτ=12∫εE2 dτW = \tfrac12\int\vec{D}\cdot\vec{E}\,d\tau = \tfrac12\int\varepsilon E^2\,d\tau (linear media). Dielectrics are pulled into regions of stronger field; a slab is drawn into a capacitor by the fringing field on its edge bound charge.