University Physics II · Capacitance and Dielectrics · 5.6
Gauss's Law in Dielectric Media
A dielectric adds bound charge, and Gauss's law counts it like any other charge. Write the displacement field and only the free charge appears on the right — a shortcut that holds as far as symmetry, linearity and geometry do.
Build the model
Connect the measurement to the mechanism.
Gauss's law does not acquire a dielectric version. ∮E·dA = Qenc/ε₀ is exact in matter, provided Qenc counts every charge inside — including the bound charge that polarization pushes to the material's surfaces. The difficulty is practical: bound charge depends on E, and E depends on bound charge, so the enclosed charge is not known before the problem is solved. The electric displacement D = ε₀E + P absorbs that circularity.
Its flux law ∮D·dA = Qfree, enc references only the charge you deposited, so for a linear, isotropic, homogeneous medium with enough symmetry you read D off the free charge and get E = D/(κε₀) in one step. D is bookkeeping, not a new field: the force on a test charge is still qE, D is not curl-free, and zero free charge does not mean zero D. Every shortcut here rests on assumptions worth naming before you use it.
- Simple definition
- In matter, Gauss's law still counts every enclosed charge, free plus bound. The displacement field D = ε₀E + P repackages that count so only free charge appears on the right: ∮D·dA = Qfree, enc.
- Example
- A parallel plate carries σf = 1.77 μC m⁻² against a κ = 2.50 dielectric. D = σf, the bound charge is σb = σf(1 − 1/κ) = 1.06 μC m⁻², and the net 0.71 μC m⁻² gives E = 8.00 × 10⁴ V m⁻¹.
Exact in matter. The dielectric does not alter the law, only what sits inside the surface.
ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²; flux in V m
Uniform P in a homogeneous slab puts all the bound charge on the faces and none in the bulk.
P in C m⁻²; n̂ points out of the dielectric
A change of variable that cancels every bound term, so only deposited charge appears.
D in C m⁻²; only free charge on the right-hand side
With κ constant the route is E = D/(κε₀): get D from free charge, then divide once.
κ and χₑ dimensionless; vacuum has χₑ = 0, κ = 1
Bound charge cancels the fraction 1 − 1/κ of the plate charge, leaving E smaller by κ.
σf = 1.77 μC m⁻², κ = 2.50 → E = 8.00 × 10⁴ V m⁻¹
Spherical symmetry gives D at once; bound charge hugging q cuts the net enclosed charge.
q = 5.00 nC, κ = 4.00, r = 30.0 mm → 1.25 × 10⁴ V m⁻¹
Gauss's law was never a vacuum law
∮E·dA = Qenc/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 = (Qfree + Qbound)/ε₀. The equation stays exact. The obstacle is that Qbound is not known in advance, because the polarization producing it is itself a response to the total field the equation is trying to find.
Where the bound charge ends up
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 σb = P·n̂; the interior carries ρb = −∇·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 σf = 1.77 μC m⁻². The slab face against the positive plate acquires σb = σf(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.
D hides the unknown you cannot compute
Define D = ε₀E + P. Then ∇·D = ε₀∇·E + ∇·P, and since ε₀∇·E = ρf + ρb while ∇·P = −ρb, the bound terms cancel and ∇·D = ρf. In integral form, ∮D·dA = Qfree, 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 = σf = 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 σb. The bound charge is still there; you just never had to name it.
Symmetry, not the D law, closes the problem
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.
One sphere, worked two ways
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 qb = −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.
The assumptions you signed for
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, ρb = −∇·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.
Change one variable at a time
Make the relationship visible.
Raise κ and watch the dashed D line refuse to move — only ε₀E inside the slab sinks, and the arrow between them, P = σb, opens up to take the difference; then shorten the fill and see the vacuum layer snap straight back to ε₀E = D.
D = σf (FREE ONLY)1.77 μC m⁻²
P = σb IN THE SLAB1.06 μC m⁻²
E INSIDE THE SLAB0.80 × 10⁵ V m⁻¹
E IN VACUUM, D/ε₀2.00 × 10⁵ V m⁻¹
Live interpretationD = σf (FREE ONLY): 1.77 μC m⁻². P = σb IN THE SLAB: 1.06 μC m⁻². E INSIDE THE SLAB: 0.80 × 10⁵ V m⁻¹. E IN VACUUM, D/ε₀: 2.00 × 10⁵ V m⁻¹
Catch the common trap
Explain before calculating.
A +5.00 nC point charge sits at the centre of a thick spherical shell of linear dielectric with κ = 4.00, inner radius 20.0 mm and outer radius 60.0 mm. What is the electric field magnitude at r = 30.0 mm, inside the dielectric?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA parallel-plate capacitor has free charge density σf = 3.54 μC m⁻² on its plates, and a linear dielectric of κ = 4.00 fills the gap completely. Find D and E in the gap and the bound surface charge density σb, then check E the long way.
- Put a pillbox with one face inside the plate metal and the other inside the dielectric. ∮D·dA = Qfree, enc counts the plate charge only, so D = σf = 3.54 μC m⁻² — κ never enters this line.
- Linear isotropic medium, so D = κε₀E: E = D/(κε₀) = (3.54 × 10⁻⁶ C m⁻²)/(4.00 × 8.854 × 10⁻¹² C² N⁻¹ m⁻²) = 1.00 × 10⁵ V m⁻¹.
- Bound sheet on each slab face: σb = σf(1 − 1/κ) = 3.54 × 0.750 = 2.66 μC m⁻², negative on the face against the positive plate.
- Now the long way, with ∮E·dA = Qenc/ε₀: that same pillbox encloses 3.54 − 2.655 = 0.885 μC m⁻², and (8.85 × 10⁻⁷)/(8.854 × 10⁻¹²) = 1.00 × 10⁵ V m⁻¹ — the same field, with the bound charge named out loud.
AnswerD = 3.54 μC m⁻², E = 1.00 × 10⁵ V m⁻¹, σb = 2.66 μC m⁻² on each face. At κ = 4.00 the bound charge cancels three-quarters of the plate charge, never all of it.
MediumPlates 2.00 mm apart carry σf = 1.06 μC m⁻². A slab of κ = 5.00 and thickness 1.20 mm lies flat against one plate, leaving 0.80 mm of vacuum. Find the field in each layer and the potential difference across the gap.
- Every pillbox you can draw between the plates encloses the same free charge, so ∮D·dA = Qfree, enc gives one value right across the gap: D = σf = 1.06 μC m⁻².
- Vacuum layer, κ = 1: E₁ = D/ε₀ = (1.06 × 10⁻⁶)/(8.854 × 10⁻¹²) = 1.197 × 10⁵ V m⁻¹.
- Slab, κ = 5.00: E₂ = D/(κε₀) = E₁/5.00 = (1.197 × 10⁵)/5.00 = 2.394 × 10⁴ V m⁻¹. The bound sheets doing that, σb = σf(1 − 1/κ) = 0.848 μC m⁻², sit on the slab faces and nowhere else.
- E is uniform within each layer, so the voltages add: ΔV = E₂t + E₁(d − t) = (2.394 × 10⁴)(1.20 × 10⁻³) + (1.197 × 10⁵)(0.80 × 10⁻³) = 28.7 + 95.8 = 124.5 V.
- Dividing the whole 2.00 mm by κ instead would give (2.394 × 10⁴)(2.00 × 10⁻³) = 47.9 V, low by a factor of 2.6: κ applies where the dielectric is, not to the gap.
AnswerD = 1.06 μC m⁻² throughout; E = 2.39 × 10⁴ V m⁻¹ in the slab and 1.20 × 10⁵ V m⁻¹ in the vacuum layer; ΔV = 124.5 V. The 0.80 mm of vacuum takes 77% of the voltage across 40% of the gap.
HardA sphere of radius 25.0 mm is uniformly polarized with P = 3.00 μC m⁻² and carries no free charge anywhere. Find E and D inside it, and the flux of D through a concentric sphere of radius 15.0 mm.
- There is no free charge, so ∮D·dA = Qfree, enc = 0 through every closed surface you can draw. That is a statement about flux, not about D itself.
- The bound charge is all on the surface, σb = P·n̂ = P cos θ, and the field it makes inside a uniformly polarized sphere is uniform: E = P/(3ε₀) = (3.00 × 10⁻⁶)/(3 × 8.854 × 10⁻¹²) = 1.13 × 10⁵ V m⁻¹, pointing against P.
- Assemble D from its definition rather than from the flux law: D = ε₀E + P = −P/3 + P = 2P/3 = 2.00 × 10⁻⁶ C m⁻², parallel to P and plainly not zero.
- Flux through the 15.0 mm sphere: D is uniform there, so as much leaves as enters and ΦD = 0 — exactly what Qfree, enc = 0 demanded, with D ≠ 0 the whole time.
- P is uniform, so ρb = −∇·P = 0 and that inner sphere encloses no bound charge either, making ΦE = 0 as well. Neither flux law hands you a field here; only the surface bound charge does.
AnswerE = 1.13 × 10⁵ V m⁻¹ antiparallel to P, D = 2.00 μC m⁻² parallel to P, and ΦD = 0 through the 15.0 mm sphere. Zero flux forces zero enclosed free charge, not zero D — the law fixes a divergence, and symmetry has to do the rest.