What is the electric displacement field

Click For Summary
SUMMARY

The electric displacement field, denoted as \mathbf{D}, is defined by the equation \mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}, where \varepsilon_0 represents the permittivity of free space, \mathbf{E} is the electric field, and \mathbf{P} is the polarization density. This vector field has units of coulombs per square meter (C/m²) and is crucial for understanding the behavior of electric fields in dielectric materials. In linear media, the relationship simplifies to \mathbf{D}= \mathbf{\varepsilon}_0 \mathbf{\varepsilon}_r\mathbf{E}, where \mathbf{\varepsilon}_r is the relative permittivity tensor. The divergence of \mathbf{D} corresponds to the free charge density, as expressed in Gauss' Law: \text{div}(\mathbf{D}) = \rho_f.

PREREQUISITES
  • Understanding of vector fields in electromagnetism
  • Familiarity with Gauss' Law and Ampère's Circuital Law
  • Knowledge of dielectric materials and their properties
  • Basic grasp of permittivity and polarization concepts
NEXT STEPS
  • Study the derivation and applications of Gauss' Law in electrostatics
  • Explore the concept of polarization in dielectric materials
  • Learn about the relationship between electric displacement field and electric field in various media
  • Investigate the role of permittivity in capacitor design and performance
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetism, particularly those focusing on dielectric materials and electric field interactions.

Messages
19,865
Reaction score
10,851
Definition/Summary

The electric displacement field in a material is defined thus,

\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}

where \varepsilon_0 is the permittivity of free space, \mathbf{E} is the electric field, and \mathbf{P} is the polarisation density of the electric material.

Electric displacement field is a vector field with units of coulombs per square metre (C/m²), and dimensions of charge/length².

Equations

The electric displacement field in linear media,
\underline{D} = \varepsilon_0\underline{E}+\underline{P}

Gauss' Law,
\text{div}\left(\underline{D}\right) = \rho_f

Ampère's Circuital Law (modified by Maxwell),
\text{curl}\left(\underline{H}\right) = \underline{j} +\frac{\partial\underline{D}}{\partial t}

At the surface of a conductor:
\underline{D} = \sigma\underline{\hat{n}}

Extended explanation

The electric displacement field is a vector field which describes the displacement effects of an electric field on the charges within a dielectric material, such as polarisation charges or bound charges.

The electric displacement field has the same units as the dielectric polarisation, dipole moment per unit volume.

In linear media the polarisation may be written:

\mathbf{P}=\left(\mathbf{\varepsilon}_r-\widetilde{\mathbf{I}}\right)\mathbf{\varepsilon}_0\mathbf{E}

where \mathbf{\varepsilon}_r is the relative permittivity tensor of the dielectric. Hence, the electric displacement field in linear materials may be written as,

\mathbf{D}= \mathbf{\varepsilon}_0 \mathbf{\varepsilon}_r\mathbf{E}

If the dielectric is isotropic, then the relative permittivity tensor is simply a numerical constant times the unit tensor, \widetilde{\mathbf{I}}, and so may be written as a number.

It should be noted at this point that unlike the electric field, the electric displacement field has no real physical meaning, it only serves to make calculations involving dielectric materials much easier.

Reason for name of electric displacement field:

Technically, only the polarisation field, \mathbf{P}, is a displacement field.

The polarisation field measures the displacement of charges within material.

At points in a vacuum, where of course there are no charges to be displaced, \mathbf{D} will generally be non-zero (it will be a multiple of \mathbf{E}), showing that at those points \mathbf{D} is entirely not a "displacement field".

\mathbf{D}, despite its name, is technically a combination of a displacement field and a non-displacement field.

Reason for definition of electric displacement field:

Electric displacement field \mathbf{D} is defined so that its divergence (at each point) is the free charge density:

\nabla\cdot\mathbf{D} = \rho_f = \frac{\partial^3q}{\partial x\partial y\partial z}

So it must have dimensions of charge/area, and, like \mathbf{P}, can be measured in coulombs/metre².

By comparison, electric field \mathbf{E} is measured in volts/metre.

At a conductor:

Inside a conductor (in equilibrium), there is no electric displacement field: \boldsymbol{D}=0

At the surface of a conductor (in equilibrium), the electric displacement field is perpendicular to the surface and has magnitude equal to the surface charge density: \boldsymbol{D}=\sigma\boldsymbol{\hat{n}}

This is because both D and ρf are discontinuous, so Gauss' law, divD = ρf, becomes D = ∆D = ∆ρf = σ

In particular, in a parallel-plate capacitor (in which the whole charge, Q, is on the inside face of each plate):

D=\sigma = Q/A, and so E=Q/\varepsilon A​

Permittivity:

The vector fields \mathbf{D} and \mathbf{E} are related by the tensor field \widetilde{\mathbf{\varepsilon}}, the permittivity tensor:

\mathbf{D} = \widetilde{\mathbf{\varepsilon}} \mathbf{E} = \mathbf{\varepsilon}_0 \widetilde{\mathbf{\varepsilon}}_r \mathbf{E}

Permittivity has dimensions of charge²/force.area, or charge².time²/mass.length³, and is measured in units of farads/metre.

By comparison, the magnetic analogy of permittivity is, for historical reasons, the inverse of permeability.

The vector fields \mathbf{H} and \mathbf{B} are related by the tensor field \widetilde{\mathbf{\mu}}^{-1}, the inverse of the permeability tensor:
\mathbf{H} = \widetilde{\mathbf{\mu}}^{-1}\mathbf{B} = \frac{1}{\mu_0}\,\widetilde{\mathbf{\mu}}_r^{-1}\mathbf{B}

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
  • Like
Likes   Reactions: sulhar and (deleted member)
Physics news on Phys.org
Very nice write up.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
840
  • · Replies 1 ·
Replies
1
Views
604
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 5 ·
Replies
5
Views
665
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
553
  • · Replies 6 ·
Replies
6
Views
902