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Derivation

Dielectrics, Bound Charge and the D Field

Statement

A dielectric with polarization field P(r) is electrostatically equivalent to a bound volume charge density ρb = −∇·P together with a bound surface charge density σb = P·n̂. Consequently the auxiliary field D ≡ ε0E + P satisfies Gauss's law sourced by the free charge alone, ∇·D = ρf, with the integral form ∮D·da = Qf,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. Separating 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 every atom.

The field 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 P = dipole moment per unit volume.If the charge distribution has no well-defined local dipole density (e.g. free charges, or scales below a molecule where "per unit volume" is meaningless) the multipole picture that produces ρb = −∇·P does not hold.
Electrostatics: charges are static, ∂P/∂t = 0.A time-varying polarization carries a polarization current Jp = ∂P/∂t; charge conservation then couples bound charge to that current, and the static bookkeeping alone is incomplete.
The potential of the dielectric is the superposition of point-dipole potentials, truncated at the dipole term.Retaining quadrupole and higher moments adds gradient-of-gradient source terms (∼∇∇:Q); dropping the assumption means ρb = −∇·P is only the leading contribution to the bound charge.
Fields and P are the macroscopic (spatially averaged) quantities, not the microscopic atomic fields.Without coarse-graining over many molecules, E and P fluctuate wildly between and inside atoms, and neither the divergence relation nor a smooth D is meaningful.
Derivation
1
V(r) = (1/4πε0) ∫𝒱′ [ P(r′)·(r−r′) / |r−r′|³ ] d³r′
Each volume element carries dipole moment dp = P d³r′; sum the point-dipole potentials (from the multipole-expansion result). A
2
(r−r′) / |r−r′|³ = ∇′ ( 1 / |r−r′| )
Gradient identity, taken with respect to the source coordinate r′ (note the sign: ∇′ of 1/|r−r′| points toward the field point). A
3
V = (1/4πε0) ∫𝒱′ P·∇′( 1/|r−r′| ) d³r′
Substitute step 2 into step 1. A
4
∇′·( P/|r−r′| ) = P·∇′(1/|r−r′|) + (1/|r−r′|) (∇′·P)
Product rule for the divergence of a scalar times a vector, used to convert the integrand into a total divergence plus a remainder. B
5
V = (1/4πε0) [ ∫𝒱′ ∇′·( P/|r−r′| ) d³r′ − ∫𝒱′ (∇′·P)/|r−r′| d³r′ ]
Rearrange step 4 for P·∇′(…) and insert into step 3. B
6
V = (1/4πε0) [ ∮𝒮′ ( P·n̂ )/|r−r′| da′ + ∫𝒱′ ( −∇′·P )/|r−r′| d³r′ ]
Divergence theorem turns the first volume integral into a surface integral over the boundary 𝒮′ with outward normal n̂. B
7
σb ≡ P·n̂ ,   ρb ≡ −∇·P
Identify each kernel 1/|r−r′| against the Coulomb potential of a surface/volume charge: whatever multiplies it is that charge density. A
8
V(r) = (1/4πε0) [ ∮𝒮′ σb/|r−r′| da′ + ∫𝒱′ ρb/|r−r′| d³r′ ]
The polarized object is exactly a set of real charges ρb, σb: bound charge is genuine charge, not a mathematical fiction. A
9
∇·E = ρ/ε0 = ( ρf + ρb )/ε0
Differential Gauss's law is sourced by all charge; split the total density into free plus bound (from the Gauss-divergence result). A
10
ε0(∇·E) = ρf − ∇·P
Multiply by ε0 and substitute ρb = −∇·P from step 7. A
11
∇·( ε0E + P ) = ρf
Bring ∇·P to the left; linearity of the divergence lets the two terms combine under one ∇·. A
12
D ≡ ε0E + P  ⇒  ∇·D = ρf
Define the displacement field as the quantity whose divergence is the free charge; the definition is what removes the bound charge from view. A
13
∫𝒱 ∇·D d³r = ∮𝒮 D·da = ∫𝒱 ρf d³r = Qf,enc
Integrate step 12 over an arbitrary volume and apply the divergence theorem to obtain the integral form. Valid even across surfaces where P jumps, since the derivation used only ∇·D=ρf in the distributional sense. C
Result
ρb = −∇·P ,   σb = P·n̂ ,   D = ε0E + P ,   ∇·D = ρf ,   ∮D·da = Qf,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, the normal component of P counts the dipole heads left stranded on the boundary. The displacement D is the specific combination of the true field E and the material's response P whose flux through any closed surface counts only the free charge you put there — the bound charge cancels internally.

Units check. [P] = C·m⁻² (dipole moment C·m per volume m³). Then [∇·P] = C·m⁻³ = charge density ✓, and [P·n̂] = C·m⁻² = surface charge density ✓. [ε0E] = (C²·N⁻¹·m⁻²)(N·C⁻¹) = C·m⁻², matching [P], so [D] = C·m⁻² ✓. [∇·D] = C·m⁻³ = [ρf] ✓.

Limiting cases
  • Uniform polarization (P = const): ∇·P = 0, so ρb = 0 and all bound charge sits on the surfaces as σb = P·n̂ — e.g. a uniformly polarized slab or sphere.
  • Vacuum / no matter (P = 0): D = ε0E and ∇·D = ρf collapses to ordinary Gauss's law.
  • Linear isotropic dielectric (P = ε0χeE): D = ε0(1+χe)E = εE, and inside a region of only free charge, ρb = −(χe/(1+χe)) ρf.
  • Conductor limit (χe → ∞, ε → ∞): E → 0 inside, bound surface charge screens the free charge completely.
Breaks when
  • Time-dependent polarization. With ∂P/∂t ≠ 0 a polarization current Jp = ∂P/∂t flows; charge conservation still holds (∇·Jp = −∂ρb/∂t), but the electrostatic potential of step 1 is no longer valid and one must use the full Maxwell equations. ∇·D = ρf survives, but ∇×E ≠ 0.
  • Strongly inhomogeneous fields (nonlocal / gradient response). When P at a point depends on E elsewhere (over atomic-scale gradients, near interfaces, in metals' Thomas–Fermi screening), the simple local dipole density fails and higher multipoles (quadrupole density) contribute additional bound charge beyond −∇·P.
  • Nonlinear / hysteretic media (ferroelectrics). If P is not a single-valued function of E, D = εE is meaningless; ∇·D = ρf still holds as bookkeeping but no longer closes the boundary-value problem without the material's P(E) history.
  • Below the coarse-graining scale. At sub-molecular resolution there is no smooth P; only discrete microscopic charges exist and the macroscopic D loses meaning.
Failure modes
  • Sign flip on ρb. Writing ρb = +∇·P. The minus sign is not cosmetic: it comes from the product rule (step 5), and getting it wrong reverses every bound-charge field.
  • Thinking D has no curl or that "D lines start on free charge only." ∇×D = ∇×P ≠ 0 in general, so D is not derivable from a potential and Gauss's law for D alone does not determine it without a constitutive relation.
  • Using D = εE in a nonlinear or anisotropic medium. The scalar ε exists only for linear isotropic materials; in crystals ε is a tensor and D need not be parallel to E.
  • Forgetting σb at interfaces. Applying ρb = −∇·P in the bulk while dropping the surface term double-counts or loses charge; total bound charge ∮σbda + ∫ρbd³r = 0 must hold.
  • Treating bound charge as fictitious. Students discount ρb as "not real," yet it produces real E fields and real forces; it is simply charge you did not place by hand.
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 −∇·P and P·n̂. 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 vs 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 (bound charge).

The field D is then a construction of convenience: we bundle the awkward −∇·P source back into the field so its divergence sees only ρf. This is enormously useful because ρf is what an experimenter specifies. But it comes at a price: D is not a fundamental field. Its curl is set by ∇×P, so unlike E it is generally not curl-free and cannot be obtained from Gauss's law plus symmetry alone except in cases of high symmetry where ∇×P = 0 by construction.

The boundary conditions that fall out — D⊥,above − D⊥,below = σf and E∥,above = E∥,below — are the workhorses of dielectric problems: the normal component of D is continuous across an uncharged interface while the tangential E is always continuous. Together with a constitutive relation they close every capacitor and interface problem, and their combination is what bends field lines at a dielectric boundary (the electrostatic analogue of refraction).

A subtlety worth stating precisely: the split into ρb = −∇·P is exact only if P captures the full leading dipole density and higher moments are negligible. In a rigorous microscopic-to-macroscopic averaging (Russakoff's derivation), the macroscopic charge density is ⟨ρmicro⟩ = ρf − ∇·P + ∇∇:Q − …, where Q is a quadrupole density. The auxiliary field is more properly Di = ε0Ei + Pi − ∂jQij + …; the familiar D = ε0E + P is the electric-dipole approximation, excellent for ordinary dielectrics but not a fundamental law.

Common misconceptions. D is not "the field in a dielectric" — that is E; D is a bookkeeping field. Gauss's law for D does not make D depend only on free charge in geometry (it depends on E and P, hence on bound charge too); only its divergence and hence flux are blind to bound charge. And "no free charge" does not mean "no D": a uniformly polarized sphere in vacuum has ρf = 0 everywhere yet nonzero D inside.

Worked examples
1
Uniformly polarized dielectric sphere. Sphere radius R = 2.0 cm, uniform P = P ẑ with P = 3.0×10⁻⁶ C·m⁻². Find ρb, σb(θ), and total bound charge on the upper hemisphere.
Set up: symbols first, then numbers. A
2
ρb = −∇·P = 0  (since P is constant)
Uniform vector field has zero divergence. A
3
σb = P·n̂ = P ẑ·r̂ = P cos θ
On a sphere the outward normal is r̂, and ẑ·r̂ = cos θ. A
4
Qupper = ∫02π∫0π/2 P cos θ · R² sin θ dθ dφ = 2πPR² ∫0π/2 cos θ sin θ dθ = πPR²
Integrate σb over the upper hemisphere; ∫0π/2cosθ sinθ dθ = 1/2. B
5
Qupper = π (3.0×10⁻⁶)(0.020)² = π (3.0×10⁻⁶)(4.0×10⁻⁴) = 3.77×10⁻⁹ C
Substitute numbers; units C·m⁻² · m² = C. A
ρb = 0,   σb = P cos θ = (3.0 μC·m⁻²) cos θ,   Qupper ≈ +3.8 nC

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

1
Free charge inside a linear dielectric. Point free charge Qf = 5.0 nC embedded in an infinite linear dielectric with εr = 1 + χe = 4.0. Find D(r), E(r), and the total bound charge qb hugging the point charge.
Set up with symbols first. A
2
∮D·da = Qf,enc ⇒ D(4πr²) = Qf ⇒ D = Qf/(4πr²) r̂
Spherical symmetry: D is radial and constant on the Gaussian sphere; only free charge sources it. A
3
E = D/ε = D/(ε0εr) = Qf/(4πε0εrr²) r̂
Linear isotropic constitutive relation D = εE. The field is reduced by 1/εr vs vacuum. A
4
qb = −(χe/εr) Qf = −((εr−1)/εr) Qf
The point bound charge equals −∇·P integrated over an infinitesimal shell; with P = ε0χeE this is the standard screening result. B
5
D = (5.0×10⁻⁹)/(4πr²) = 3.98×10⁻¹⁰/r² C·m⁻²;   qb = −(3.0/4.0)(5.0 nC) = −3.75 nC
Insert numbers; at r = 1 cm, E = 3.98×10⁻¹⁰/(8.85×10⁻¹²·4·10⁻⁴) ≈ 1.12×10⁴ N·C⁻¹. A
D = Qf/(4πr²),   E = D/(ε0εr),   qb = −3.75 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 Qf/εr = 1.25 nC, exactly the 1/4 reduction of E. Units check: [D] = C·m⁻² ✓, [E] = C·m⁻²/(C²N⁻¹m⁻²) = N·C⁻¹ ✓.

Problems
  1. (A) A slab of dielectric fills 0 ≤ z ≤ d with polarization P = P0(z/d) ẑ. Find ρb and the two surface densities.
    Solutionρb = −∇·P = −∂z(P0 z/d) = −P0/d (uniform, negative). At z=d (top, n̂=+ẑ): σb = P·n̂ = P0. At z=0 (bottom, n̂=−ẑ): P(0)=0 so σb=0. Total per area: −(P0/d)·d + P0 = 0 ✓ (neutral).
  2. (A) A dielectric sphere (radius R, permittivity ε) contains uniform free charge density ρf. Find D and E inside.
    SolutionBy symmetry ∮D·da = Qf,enc: D(4πr²) = ρf(4/3)πr³ ⇒ D = ρf r/3 (radial), for r≤R. Then E = D/ε = ρf r/(3ε). Note D is independent of ε because only free charge and geometry enter Gauss's law for D.
  3. (B) Show explicitly that the total bound charge of any finite polarized body is zero, i.e. ∫𝒱 ρb d³r + ∮𝒮 σb da = 0.
    Solution∫ρbd³r = −∫∇·P d³r = −∮P·da (divergence theorem). ∮σbda = ∮P·n̂ da = +∮P·da. The two are equal and opposite, so the sum vanishes. Physically: polarization merely rearranges internal charge, creating no net charge.
  4. (B) A parallel-plate capacitor, plate area A = 100 cm², gap d = 1.0 mm, holds free surface charge σf = 2.0×10⁻⁶ C·m⁻² and is filled with a dielectric εr = 5.0. Find D, E, P, σb, and the voltage.
    SolutionD = σf = 2.0×10⁻⁶ C·m⁻² (from ∮D·da=Qf,enc across a pillbox). E = D/(ε0εr) = 2.0×10⁻⁶/(8.85×10⁻¹²·5.0) = 4.52×10⁴ V·m⁻¹. P = D − ε0E = 2.0×10⁻⁶ − 8.85×10⁻¹²·4.52×10⁴ = 1.60×10⁻⁶ C·m⁻². σb = P·n̂ = 1.60×10⁻⁶ C·m⁻² (opposite sign to adjacent free charge). Voltage V = Ed = 4.52×10⁴·1.0×10⁻³ = 45.2 V. Check: σf−σb = 0.40×10⁻⁶ = σf/εr ✓.
  5. (C) A long cylinder (radius a) carries "frozen-in" polarization P = k s ŝ (radial, s the cylindrical radius, k constant), no free charge. Find E inside using bound charge, then verify with D.
    Solutionρb = −∇·P = −(1/s)∂s(s·ks) = −(1/s)(2ks) = −2k. Surface σb = P(a)·ŝ = ka. Gauss (Gaussian cylinder radius s<a, length L): E(2πsL) = ρb(πs²L)/ε0 ⇒ E = ρbs/(2ε0) = −ks/ε0 (radial, inward). Verify with D: no free charge and symmetry give D = 0 everywhere; then E = (D−P)/ε0 = −P/ε0 = −ks/ε0 ŝ ✓. This example warns that D=0 does not mean E=0.