Polarisation, magnetisation and material fields
| English | Português |
|---|---|
| electric polarisation | electric polarisation |
| magnetisation | magnetisation |
A decision before an answer
- Putting a dielectric into a capacitor changes the field differently depending on whether its battery remains connected.
- Your goal: Relate free and bound charge to D, E and polarisation.
Separate free and bound charge
- Electric polarisation P is electric dipole moment per unit volume, measured in C/m². Define D=ε₀E+P so that ∇·D=ρ_free; the total charge, including bound charge, still appears in ∇·E=ρ_total/ε₀. Bound volume charge is ρ_b=−∇·P and bound surface charge is σ_b=P·n, where n points outward from the material.
- Uniform P therefore gives no bound charge in the bulk but can give surface charges of opposite sign. A dielectric is not an ideal metal: polarisation need not make its internal electric field zero.
An isolated ideal parallel-plate capacitor is fully filled with a linear dielectric of ε_r=3. Which pair gives voltage and stored-energy ratios?
Free Q stays fixed and C triples; both V=Q/C and U=Q²/(2C) become one third.
Apply the material relation
- For a linear isotropic dielectric, P=ε₀χ_e E and D=εE with ε=ε₀(1+χ_e)=ε₀ε_r. These simple scalar relations assume the response is linear and ignore anisotropy, strong dispersion and nonlinear effects.
- In a fully filled, large parallel-plate capacitor with negligible edge effects, D normal to the plates equals the free surface-charge density. If σ_free stays fixed, increasing ε reduces E=σ_free/ε. The bound charges oppose the applied field; their magnitude is not automatically equal to the free plate charge.
Uniform electric polarisation in a finite dielectric necessarily implies:
ρ_b=−∇·P=0 for uniform P, but σ_b=P·n can be nonzero.
Hold charge or voltage fixed
- For plate area A and spacing d, C=εA/d. Disconnecting the battery fixes free charge Q: V=Q/C falls and stored energy U=Q²/(2C) falls when a dielectric increases C. Leaving an ideal voltage source connected fixes V: Q=CV and stored energy U=CV²/2 both rise.
- The source supplies energy and moving the dielectric can involve mechanical work; compare the stated electrical energy quantity rather than assuming the capacitor alone is an isolated system. At a material interface n·(D₂−D₁)=σ_free, whereas static tangential E is continuous. Normal E generally changes when permittivity changes.
Take a fully filled capacitor with ε_r=4 and fixed free plate density σ. Gauss’s law for D gives D=σ, then E=σ/(4ε₀) and P=ε₀(4−1)E=3σ/4. C rises fourfold, so fixed-Q voltage and stored energy become one quarter. With the battery connected instead, V and E=V/d stay fixed while Q and U become four times larger. Separately, H=400 A/m and χ_m=0.5 give M=200 A/m and B=μ₀(600 A/m)≈0.754 mT.
In a linear magnetic material with H=500 A/m and χ_m=0.2, M=____ A/m.
M=χ_mH=0.2·500=100.
Distinguish magnetic fields
- Magnetisation M is magnetic dipole moment per unit volume, measured in A/m. In SI, B=μ₀(H+M); for a linear isotropic response M=χ_m H and B=μ₀(1+χ_m)H. B and H have different units and roles.
- A long uniform solenoid has H≈nI when end and demagnetising effects are negligible; material response then changes B. Free surface current sets n×(H₂−H₁)=K_free, and normal B remains continuous. Ferromagnetic hysteresis and saturation cannot be represented by one constant χ_m over every field.
Use free charge in the D equation and total charge in the E equation. Do not confuse fixed Q with fixed V, or identify B numerically with H.
Which answer fits this case?
Relate free and bound charge to D, E and polarisation
Normal electric field must be continuous across every dielectric interface even if the permittivities differ.
With no free surface charge, normal D is continuous; E=D/ε can differ.
Keep the distinctions
- electric polarisation 电极化强度 — Electric dipole moment per unit volume, P, contributing bound charge.
- magnetisation 磁化强度 — Magnetic dipole moment per unit volume, M, entering B=μ₀(H+M).
- Relate free and bound charge to D, E and polarisation.
- Compare fixed-charge and fixed-voltage dielectric changes.
- Use magnetic constitutive response and free-current boundary conditions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.