1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the electric displacement field

  1. Jul 24, 2014 #1
    Definition/Summary

    The electric displacement field in a material is defined thus,

    [tex]\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}[/tex]

    where [itex]\varepsilon_0[/itex] is the permittivity of free space, [itex]\mathbf{E}[/itex] is the electric field, and [itex]\mathbf{P}[/itex] 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,
    [tex]\underline{D} = \varepsilon_0\underline{E}+\underline{P}[/tex]

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

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

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

    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:

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

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

    [tex]\mathbf{D}= \mathbf{\varepsilon}_0 \mathbf{\varepsilon}_r\mathbf{E}[/tex]

    If the dielectric is isotropic, then the relative permittivity tensor is simply a numerical constant times the unit tensor, [tex]\widetilde{\mathbf{I}}[/tex], 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, [itex]\mathbf{P}[/itex], 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, [itex]\mathbf{D}[/itex] will generally be non-zero (it will be a multiple of [itex]\mathbf{E}[/itex]), showing that at those points [itex]\mathbf{D}[/itex] is entirely not a "displacement field".

    [itex]\mathbf{D}[/itex], 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 [itex]\mathbf{D}[/itex] is defined so that its divergence (at each point) is the free charge density:

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

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

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

    At a conductor:

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

    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: [itex]\boldsymbol{D}=\sigma\boldsymbol{\hat{n}}[/itex]

    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):

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


    Permittivity:

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

    [tex]\mathbf{D} = \widetilde{\mathbf{\varepsilon}} \mathbf{E} = \mathbf{\varepsilon}_0 \widetilde{\mathbf{\varepsilon}}_r \mathbf{E}[/tex]

    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 [itex]\mathbf{H}[/itex] and [itex]\mathbf{B}[/itex] are related by the tensor field [itex]\widetilde{\mathbf{\mu}}^{-1}[/itex], the inverse of the permeability tensor:

    [tex]\mathbf{H} = \widetilde{\mathbf{\mu}}^{-1}\mathbf{B} = \frac{1}{\mu_0}\,\widetilde{\mathbf{\mu}}_r^{-1}\mathbf{B}[/tex]

    * 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!
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted