University Physics II · Capacitance and Dielectrics · 5.4
Dielectric Polarization
Charge that cannot leave its molecule can still shift. That shift leaves bound sheets on the surfaces, those sheets oppose the applied field, and the strength of the opposition is what the dielectric constant measures.
Build the model
Connect the measurement to the mechanism.
A dielectric has no charge free to travel, but its charge can still shift. An applied field displaces every electron cloud slightly against its nucleus, and in a polar material it also biases the tumbling of permanent dipoles. Both give each small volume a dipole moment per unit volume, P, in C m⁻².
Where P meets a surface it leaves uncompensated bound charge, σb = P·n̂, and that sheet makes a field opposing the one that created it. Hence the loop that defines the subject: the applied field polarizes the material, the polarization weakens the field, and the weakened field is what sets the polarization. For a linear material, P = ε₀χₑ E closes the loop at a fixed point, E = E₀/(1 + χₑ) = E₀/K.
So the dielectric constant is not an extra ingredient; it is the answer to a self-consistency condition, and the susceptibility χₑ counts how strongly the material talks back.
- Simple definition
- Polarization is the displacement or alignment of a material's bound charge under an applied field, measured by P, the dipole moment per unit volume in C m⁻². Its surfaces then carry bound charge that partly cancels the applied field.
- Example
- Fill a capacitor gap with a K = 4.0 slab while the plates hold σf = 8.85 × 10⁻⁶ C m⁻². Bound faces of 6.64 × 10⁻⁶ C m⁻² appear, and the field falls from 1.00 × 10⁶ to 2.50 × 10⁵ V m⁻¹.
Uniform P in a homogeneous slab gives ρb = 0, so all the bound charge sits on the two faces.
P and σb in C m⁻²; ρb in C m⁻³
χₑ = 3.0 for a K = 4.0 slab; vacuum has χₑ = 0. ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
χₑ dimensionless; E is the field inside the material, not the applied field
The polarization field subtracts, so solving for E returns the applied field divided by K.
Slab filling the gap, P normal to the faces, free plate charge held fixed
K ≥ 1 always. A conductor is the K → ∞ limit, with the interior field driven to zero.
Air 1.00059, paper ≈ 3.5, water 80 at 20 °C and 55 at 100 °C
R = 1.0 × 10⁻¹⁰ m and E = 1.0 × 10⁶ V m⁻¹ give p = 1.1 × 10⁻³⁴ C m, a shift of 6.9 × 10⁻¹⁶ m.
α in C m² V⁻¹; R the atomic radius, of order 10⁻¹⁰ m
Alignment competes with thermal tumbling, so this part of χₑ falls as 1/T. Induced α does not.
Valid while pE ≪ kT; k = 1.381 × 10⁻²³ J K⁻¹, n in m⁻³
Bound charge cannot leave, but it can shift
An insulator has no charge free to cross the sample, so a field cannot drive a steady current through it. What a field can do is move charge a short way inside each molecule, and stop. Two mechanisms do this. Induced polarization shifts an atom's electron cloud against its nucleus; it operates in every material, hydrogen through polyethylene, and it vanishes the instant the field is removed. Orientational polarization acts only on molecules that are already dipoles — water, HCl, plastics with polar side groups — and biases their random tumbling so that slightly more point along E than against it. Both leave the same signature: each small volume acquires a net dipole moment while every molecule in it stays neutral. The sample as a whole stays neutral too. Nothing has been added or removed; charge has only been redistributed, and by remarkably small distances.
Induced polarization: the electron cloud leans
Model a neutral atom as a nucleus of charge +q sitting inside a uniform electron cloud of radius R. An applied field pulls the two apart until the internal Coulomb attraction balances it, which for this model happens at a dipole moment p = αE with α = 4πε₀R³ — the polarizability, fixed by the atom's volume. Take R = 1.0 × 10⁻¹⁰ m and E = 1.0 × 10⁶ V m⁻¹, about a third of the breakdown field of dry air. Then α = 1.11 × 10⁻⁴⁰ C m² V⁻¹ and p = 1.11 × 10⁻³⁴ C m; dividing by e gives a displacement of 6.9 × 10⁻¹⁶ m, seven millionths of the atomic radius. That tininess is why the response stays linear across any laboratory field: the restoring force never leaves its Hooke's-law region. It is also why α, and the part of χₑ it produces, barely depends on temperature — nothing here is competing with thermal energy.
Orientational polarization: a bias, not a parade
A polar molecule already carries a dipole moment; water's is 6.2 × 10⁻³⁰ C m. A field applies the torque τ = p × E that would align it, thermal collisions keep knocking it out of line, and Boltzmann statistics settle the contest between the alignment energy pE and the thermal energy kT. While pE ≪ kT the mean component along the field is ⟨pz⟩ = p²E/(3kT), linear in E and falling as 1/T: warm a polar dielectric and it responds less. For water at 300 K in E = 1.0 × 10⁶ V m⁻¹, pE = 6.2 × 10⁻²⁴ J against kT = 4.14 × 10⁻²¹ J, so ⟨cos θ⟩ = 5.0 × 10⁻⁴. Polarization is a statistical bias, not a parade. The same model predicts χₑ = np²/(3ε₀kT); with n = 3.35 × 10²⁸ m⁻³ for liquid water that returns χₑ ≈ 12, well short of the measured 79, because each molecule sits in its neighbours' fields as well as the applied one.
Polarization becomes surface charge
Add the molecular dipole moments in a small volume and divide by it: P = (Σ p)/V is the polarization, in C m⁻². Inside a uniformly polarized slab nothing accumulates, because each molecule's positive end sits against its neighbour's negative end — but at the surfaces the cancellation runs out and leaves bound sheets, positive where P points out of the material and negative where it points in. The general statements are σb = P·n̂ and ρb = −∇·P, so bound charge fills a volume only where P varies from place to place. Take a slab filling a capacitor gap whose plates hold σf = 8.85 × 10⁻⁶ C m⁻², giving E₀ = σf/ε₀ = 1.00 × 10⁶ V m⁻¹; with K = 4.0 the faces carry 6.64 × 10⁻⁶ C m⁻². These charges are bound in the strict sense — earth a face and nothing flows, because they are molecular ends, not carriers.
The field settles where the response stops changing it
The two bound sheets face each other across the slab exactly as the capacitor plates do, so together they make a uniform field of magnitude P/ε₀ pointing against E₀. What a molecule actually feels is therefore E = E₀ − P/ε₀, smaller than what you applied. But P was set by that reduced field, not by E₀. Substitute the linear law P = ε₀χₑ E: E = E₀ − χₑ E, so E(1 + χₑ) = E₀ and E = E₀/(1 + χₑ). Define K = 1 + χₑ and the result is E = E₀/K. This is the point of the lesson. K is not an extra ingredient bolted on afterwards; it is the solution of a feedback loop, and χₑ measures how hard the material argues back. χₑ = 0 is vacuum. χₑ = 3.0 gives K = 4.0, a field cut to a quarter and bound faces of σf(1 − 1/K) = 0.75σf. Push χₑ → ∞ and E → 0 with σb → σf: that is a conductor.
Where linear, isotropic and homogeneous give out
Four assumptions hide inside P = ε₀χₑ E. Linear: χₑ independent of E. It survives because saturation needs pE comparable to kT, which for water means E ≈ kT/p = 6.7 × 10⁸ V m⁻¹, while dry air breaks down at 3 × 10⁶ V m⁻¹ and the toughest polymer films fail well below 10⁹ — a dielectric arcs long before its dipoles come close to lining up. Isotropic: P parallel to E, which fails in a crystal such as calcite, where χₑ is a tensor and P leans off-axis. Homogeneous: one χₑ throughout, which fails at interfaces and in composites. Static: χₑ independent of frequency. Molecular rotation in water takes of order 10⁻¹¹ s, so its K falls from 80 at low frequency to n² = 1.77 in visible light, where only the electron clouds keep up. Ferroelectrics abandon linearity outright, with K in the thousands and a hysteresis loop.
Change one variable at a time
Make the relationship visible.
Drag χₑ up from zero and watch the bound sheets thicken as the net arrow shrinks, always leaving a gap: σb = σf(1 − 1/K) never catches σf, and E = E₀/K never reaches zero. Then move σf and check both gaps hold the same fraction — K belongs to the material, not to how hard you drive it.
DIELECTRIC CONSTANT K4.0
FIELD IN SLAB E0.25 MV m⁻¹
BOUND SHEET σb = P6.64 μC m⁻²
σb / σf0.750
Live interpretationDIELECTRIC CONSTANT K: 4.0. FIELD IN SLAB E: 0.25 MV m⁻¹. BOUND SHEET σb = P: 6.64 μC m⁻². σb / σf: 0.750
Catch the common trap
Explain before calculating.
A parallel-plate capacitor is charged and then disconnected, leaving a fixed free charge density σf = 8.85 × 10⁻⁶ C m⁻² on the plates. A slab of dielectric constant K = 4.0 is slid in until it fills the gap. What bound surface charge density appears on each slab face, and what is the field inside the slab?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA polythene slab (K = 2.3) is slid into the gap of a charged, disconnected parallel-plate capacitor whose plates carry a free charge density σf = 1.77 × 10⁻⁵ C m⁻². Find the applied field, the field inside the slab, and the bound charge density on each face. Take ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
- Disconnected means σf is fixed, so the plates alone set the applied field: E₀ = σf/ε₀ = 1.77 × 10⁻⁵ ÷ 8.85 × 10⁻¹² = 2.00 × 10⁶ V m⁻¹.
- The self-consistency result needs no further input: E = E₀/K = 2.00 × 10⁶ ÷ 2.3 = 8.70 × 10⁵ V m⁻¹.
- The bound sheets cancel the fraction 1 − 1/K = 1 − 0.4348 = 0.5652 of the free sheet: σb = 0.5652 × 1.77 × 10⁻⁵ = 1.00 × 10⁻⁵ C m⁻², negative on the face pressed against the positive plate.
- Check it the other way. χₑ = K − 1 = 1.3, so P = ε₀χₑE = 8.85 × 10⁻¹² × 1.3 × 8.70 × 10⁵ = 1.00 × 10⁻⁵ C m⁻² — the same number, because P is normal to the faces and σb = P·n̂.
AnswerE₀ = 2.00 × 10⁶ V m⁻¹, E = 8.70 × 10⁵ V m⁻¹, and σb = P = 1.00 × 10⁻⁵ C m⁻² on each face.
MediumA slab is slid into an applied field E₀ = 5.00 × 10⁵ V m⁻¹ held fixed by the free plate charge, and a probe reads 1.25 × 10⁵ V m⁻¹ inside it. Find K and χₑ, the polarization P, the bound charge density, and the fraction of the free charge the bound sheets cancel.
- The measurement is the ratio E = E₀/K rearranged: K = E₀/E = 5.00 × 10⁵ ÷ 1.25 × 10⁵ = 4.00, so χₑ = K − 1 = 3.00.
- The depolarising field is exactly what went missing: P/ε₀ = E₀ − E = 3.75 × 10⁵ V m⁻¹, so P = 8.85 × 10⁻¹² × 3.75 × 10⁵ = 3.32 × 10⁻⁶ C m⁻².
- Uniform P means no bound charge in the bulk, so all of it lands on the faces: σb = P = 3.32 × 10⁻⁶ C m⁻².
- The free sheets carry σf = ε₀E₀ = 8.85 × 10⁻¹² × 5.00 × 10⁵ = 4.43 × 10⁻⁶ C m⁻², and σb/σf = (E₀ − E)/E₀ = 0.750 — exactly 1 − 1/K. Three-quarters cancelled, and never all of it while K is finite.
AnswerK = 4.00 and χₑ = 3.00; P = σb = 3.32 × 10⁻⁶ C m⁻², cancelling 0.750 of the free density σf = 4.43 × 10⁻⁶ C m⁻².
HardA water molecule carries a permanent dipole moment p = 6.2 × 10⁻³⁰ C m, and liquid water holds n = 3.35 × 10²⁸ of them per cubic metre. For an internal field E = 1.0 × 10⁶ V m⁻¹ at T = 300 K, find the mean alignment ⟨cos θ⟩, the polarization P, and the susceptibility this predicts. Compare with the measured χₑ = 79. Take k = 1.381 × 10⁻²³ J K⁻¹.
- Compare the two energies before trusting the linear formula. Alignment energy pE = 6.2 × 10⁻³⁰ × 1.0 × 10⁶ = 6.2 × 10⁻²⁴ J; thermal energy kT = 1.381 × 10⁻²³ × 300 = 4.14 × 10⁻²¹ J. kT is larger by a factor of 668, so pE ≪ kT holds comfortably.
- ⟨cos θ⟩ = pE/(3kT) = 6.2 × 10⁻²⁴ ÷ (3 × 4.14 × 10⁻²¹) = 5.0 × 10⁻⁴. Five parts in ten thousand of alignment — a statistical bias, not a parade.
- ⟨pz⟩ = p⟨cos θ⟩ = 6.2 × 10⁻³⁰ × 5.0 × 10⁻⁴ = 3.09 × 10⁻³³ C m, so P = n⟨pz⟩ = 3.35 × 10²⁸ × 3.09 × 10⁻³³ = 1.04 × 10⁻⁴ C m⁻².
- χₑ = P/(ε₀E) = 1.04 × 10⁻⁴ ÷ (8.85 × 10⁻¹² × 1.0 × 10⁶) = 1.04 × 10⁻⁴ ÷ 8.85 × 10⁻⁶ = 11.7. The measured 79 is 6.8 times bigger, because each molecule sits in its neighbours' polarization field as well as the applied one.
Answer⟨cos θ⟩ = 5.0 × 10⁻⁴, P = 1.04 × 10⁻⁴ C m⁻², χₑ ≈ 12 — roughly a seventh of the measured 79, the shortfall being the neighbours' field this model ignores.