Gauss's law in dielectric media

∮E⋅dA = Q_enc/ε₀ holds in matter exactly as it holds in vacuum. Nothing in its derivation assumed empty space; it assumed that every charge inside the surface is counted. What a dielectric supplies is more charge to count. Polarization displaces bound charge — electrons shifted within atoms, permanent dipoles turned into partial alignment — and wherever that displacement is non-uniform or meets a surface, a net charge appears. It is ordinary charge: it makes fields, it feels forces, it simply cannot leave the molecule hosting it. So split the enclosure into what you deposited and what the material contributed: ∮E⋅dA = (Q_free + Q_bound)/ε₀. The equation stays exact. The obstacle is that Q_bound is not known in advance, because the polarization producing it is itself a response to the total field the equation is trying to find.

Polarization P is dipole moment per unit volume, measured in C m⁻², and it locates its own charge. A surface with outward normal n̂ carries σ_b = P⋅n̂; the interior carries ρ_b = −∇⋅P. In a homogeneous linear dielectric with no free charge sprinkled through it, ∇⋅P vanishes, so all bound charge sits on boundaries and the interior stays neutral. Take a parallel-plate gap fully filled by a κ = 2.50 slab with plates at σ_f = 1.77 μC m⁻². The slab face against the positive plate acquires σ_b = σ_f(1 − 1/κ) = 1.77 × 0.600 = 1.06 μC m⁻² of negative bound charge, and the far face an equal positive amount. A pillbox straddling the plate and that face encloses 1.77 − 1.06 = 0.71 μC m⁻², so E = (7.08 × 10⁻⁷)/(8.854 × 10⁻¹²) = 8.00 × 10⁴ V m⁻¹ — precisely the vacuum value 2.00 × 10⁵ V m⁻¹ divided by 2.50. The slab's total bound charge is zero; it has only been separated.

Define D = ε₀E + P. Then ∇⋅D = ε₀∇⋅E + ∇⋅P, and since ε₀∇⋅E = ρ_f + ρ_b while ∇⋅P = −ρ_b, the bound terms cancel and ∇⋅D = ρ_f. In integral form, ∮D⋅dA = Q_free, enc. This is a change of variable, not new physics: no instrument measures D directly, and the force on a test charge is still qE. Its value is that the right-hand side now holds only the charge you controlled. For a linear isotropic medium P = ε₀χₑ E, so D = ε₀(1 + χₑ)E = κε₀E, and the route becomes: get D from free charge plus symmetry, then divide by κε₀. In the filled capacitor a pillbox gives D = σ_f = 1.77 μC m⁻² immediately, and E = D/(κε₀) = (1.77 × 10⁻⁶)/(2.50 × 8.854 × 10⁻¹²) = 8.00 × 10⁴ V m⁻¹ — the same answer without ever computing σ_b. The bound charge is still there; you just never had to name it.

A flux law fixes a field only when symmetry lets you pull that field out of the integral, and the D law is weaker here than the E law. In electrostatics ∇×E = 0, but ∇×D = ∇×P, which is generally not zero. So any symmetry argument has to cover the polarization as well as the free charge, and the two must share the same symmetry. The clean counterexample is a uniformly polarized sphere with no free charge anywhere. Inside it E = −P/(3ε₀), so D = ε₀E + P = 2P/3, plainly not zero — yet ∮D⋅dA = 0 through every closed surface you can draw, exactly as the law demands, because there is no free charge anywhere for a surface to enclose — not because D is uniform, which it is only inside the sphere. No free charge does not mean no D. The D law is one equation; you still need symmetry, or boundary conditions, to turn it into a field.

Put q = +5.00 nC at the centre of a thick shell of κ = 4.00 dielectric, inner radius 20.0 mm and outer radius 60.0 mm, and ask for E at r = 30.0 mm. Free charge and geometry are both spherically symmetric, so D is radial and uniform on that sphere: D = q/(4πr²) = (5.00 × 10⁻⁹)/(4π × 9.00 × 10⁻⁴) = 4.42 × 10⁻⁷ C m⁻², and E = D/(κε₀) = 1.25 × 10⁴ V m⁻¹. Now the same result from bound charge alone. The shell's inner surface carries q_b = −q(1 − 1/κ) = −3.75 nC hugging the point charge, so a Gaussian sphere at 30.0 mm encloses 5.00 − 3.75 = 1.25 nC and gives E = (8.99 × 10⁹)(1.25 × 10⁻⁹)/(0.0300)² = 1.25 × 10⁴ V m⁻¹. The outer surface carries +3.75 nC, so beyond 60.0 mm the enclosed total is 5.00 nC again and the field recovers its vacuum value.

Four conditions carry the shortcut, and each fails somewhere real. Linearity: P ∝ E holds for ordinary insulators at moderate fields, but ferroelectrics such as barium titanate show hysteresis, and κ drifts with field strength as breakdown approaches. Isotropy: in calcite or a stretched polymer, D and E point in different directions and κ becomes a tensor. Homogeneity: if κ varies with position, ρ_b = −∇⋅P is non-zero in the bulk and the interior is no longer neutral. Geometry: κ applies only where the dielectric actually is, so a slab filling half the gap leaves the vacuum half at E = D/ε₀, and dividing the whole gap by κ is simply wrong. Frequency matters too — the κ describing a 50 Hz mains capacitor is not the κ at optical frequencies, where orientational polarization can no longer keep up.