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advanced27 min studyLesson 239 in Physics

Let shape determine the depolarizing field

Connect bound surface charge with internal fields and dielectric self-consistency.

Before you begin
On this page

The question for today

What you’ll understand

  • Compute bound charge from a prescribed polarization.
  • Explain why depolarization depends on sample geometry.
  • Solve the internal field of a linear dielectric sphere in a uniform applied field.

Uniform polarization does not mean that a finite object creates no electric field. Its volume bound charge may vanish while its surface bound charge remains nonzero. Those surface charges often create an internal field opposing the polarization: the depolarizing field. Its strength depends on the shape, which is why identical material in a thin slab and a sphere can respond differently to the same externally imposed field.

A dipole density produces surface charge

Polarization is electric dipole moment per unit volume. Converting its potential integral into charge form gives a volume density equal to minus its divergence and a surface density equal to its outward normal component. A spatially uniform polarization has zero divergence in the interior, but its termination at the boundary remains essential. Integrating volume and surface bound charges together yields zero net bound charge for an isolated finite polarized body.

For a sphere polarized along positive z, the surface density is proportional to cosθ. The north side is positive and the south side negative; the equator carries zero bound surface density. This is an angular degree-one pattern, so the exterior potential is dipolar and the interior potential is linear in z. Both are determined by matching continuous potential and the correct normal electric-field jump at the spherical surface, not by setting the interior field to zero.

ρb=−∇⋅P,σb=P⋅n,σbsphere=Pcos⁡θ\rho_b=-\nabla\cdot\mathbf P,\qquad \sigma_b=\mathbf P\cdot\mathbf n,\qquad \sigma_b^{\rm sphere}=P\cos\theta
P is prescribed uniform polarization magnitude in C/m², ρb is bound volume charge in C/m³, and σb is bound surface charge in C/m². n points out of the material.↔ Scroll wide equations horizontally.

Geometry selects the opposing field

For uniform polarization in an ellipsoid in vacuum, the internal depolarizing field is uniform and proportional to polarization along each principal axis. The dimensionless depolarization factors sum to one. A sphere has factor one third along every axis. A very thin broad slab polarized normally approaches factor one, while a very long needle polarized along its length approaches zero. These are shape limits under a uniform-polarization model, not universal properties of every finite object.

A nonspherical body can have a nonuniform internal field even when its imposed polarization is uniform. Edges and corners spoil the simple ellipsoid formula, and nearby conductors or a dielectric surrounding medium alter the boundary problem. It is therefore misleading to treat the factor one third as a material susceptibility. It arises from spherical geometry in vacuum; susceptibility describes the local material relation between polarization and the total internal electric field.

Edep,i=−NiPiϵ0,Nx+Ny+Nz=1,Nisphere=13E_{{\rm dep},i}=-\frac{N_iP_i}{\epsilon_0},\qquad N_x+N_y+N_z=1,\qquad N_i^{\rm sphere}=\frac13
Ni are dimensionless principal-axis depolarization factors for a uniformly polarized ellipsoid in vacuum. Edep is in V/m. This result does not assume conducting screening.↔ Scroll wide equations horizontally.

Close the material response self-consistently

For a homogeneous isotropic linear dielectric sphere, polarization is induced by the total internal field, not by the remote applied field alone. Add the sphere’s depolarizing field to the externally supplied uniform field, then substitute the constitutive law. Solving the resulting scalar relation reduces the internal field by a geometry-dependent denominator. This reduction is already a complete macroscopic boundary solution for the ideal sphere; it should not be multiplied by another ad hoc screening factor.

Use the vector lab to predict surface-charge signs from a projection. Reset and set angle to 0° and magnitude to 1; with +x as a chosen surface normal, predict positive projection. Reset and set angle to 180° and magnitude to 1; predict a negative projection. Opposite normals on a uniformly polarized body likewise carry opposite bound signs. The lab is a single vector projection, with no spherical surface or self-consistent depolarizing-field calculation.

P=ϵ0χeEin,Ein=E01+χe/3=3E0ϵr+2,ϵr=1+χe\mathbf P=\epsilon_0\chi_e\mathbf E_{\rm in},\qquad \mathbf E_{\rm in}=\frac{\mathbf E_0}{1+\chi_e/3}=\frac{3\mathbf E_0}{\epsilon_r+2},\qquad \epsilon_r=1+\chi_e
E₀ is the uniform applied vacuum field in the sphere’s absence, Ein is the total internal field, and χe is dimensionless linear susceptibility. The finite sphere is in vacuum with no free surface charge.↔ Scroll wide equations horizontally.

Work it through

A dielectric sphere screens itself

A linear dielectric sphere in vacuum has relative permittivity 4 and is placed in a remote uniform field of 60 V/m along +z. Find its internal field, polarization, depolarizing field, and north-pole bound surface charge.

  1. The susceptibility is εr−1=3. The spherical denominator is 1+χe/3=2, so the total internal field is 30 V/m along +z. It is neither the original 60 V/m nor zero.

  2. The induced polarization is ε₀×3×30=90ε₀ V/m, approximately 7.97×10⁻¹⁰ C/m² along +z. Its depolarizing field is −P/(3ε₀)=−30 V/m, making E₀+Edep=30 V/m as required.

  3. At the north pole the outward normal aligns with polarization, so σb=+P≈7.97×10⁻¹⁰ C/m². At the south pole it is the negative of this value. The integrated surface charge is zero even though the dipole moment is nonzero.

The result

The internal field is +30 V/m, polarization is +7.97×10⁻¹⁰ C/m² along z, and depolarization contributes −30 V/m. The north-pole bound charge is positive.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Predict the sign of a surface projection

Reset and set angle=0°, magnitude=1; predict x=+1. Reset and set angle=180°, magnitude=1; predict x=−1. Interpret x as polarization projected on a chosen normal.

Predict the sign of a surface projectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe projection helps track bound-charge signs. It does not integrate a spherical charge sheet or calculate a geometry-dependent depolarizing field.

Pause & reason

Check your understanding

Why does a uniformly polarized sphere have a nonzero depolarizing field despite zero interior divergence of P?

Your turn

Try it yourself

Keep χe=3 and the applied field 60 V/m, but take the ideal limit of a broad thin dielectric slab polarized normally in vacuum. Find the internal field and polarization using N=1.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve Ein=E₀−P/ε₀ together with P=ε₀χeEin.

Compare with the solution

The denominator is 1+χe=4, giving Ein=15 V/m and P=45ε₀ V/m≈3.98×10⁻¹⁰ C/m². The depolarizing field is −45 V/m. This differs from the sphere because the normal slab factor is one rather than one third; finite edges would require a more detailed boundary solution.

Keep the idea

Take it with you

  • Uniform polarization can have surface bound charge with no volume bound charge.
  • Depolarization factors describe geometry and surroundings.
  • Induced polarization responds to the total internal field self-consistently.

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