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Electric displacement field

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Vector field related to displacement current and flux density

Definition

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The electric displacement field "D" is defined as $\mathbf {D} \equiv \varepsilon _{0}\mathbf {E} +\mathbf {P} ,$ where $\varepsilon _{0}$ is thevacuum permittivity (also called permittivity of free space),E is theelectric field, andP is the (macroscopic) density of the permanent and induced electric dipole moments in the material, called thepolarization density.

The displacement field satisfiesGauss's law in a dielectric: $\nabla \cdot \mathbf {D} =\rho -\rho _{\text{b}}=\rho _{\text{f}}$

In this equation, $\rho _{\text{f}}$ is the number of free charges per unit volume. These charges are the ones that have made the volume non-neutral, and they are sometimes referred to as thespace charge. This equation says, in effect, that the flux lines ofD must begin and end on the free charges. In contrast $\rho _{\text{b}}$ , which is called the bound charge, is an effective density of the charges that are part of adipole. In the example of an insulating dielectric between metal capacitor plates, the only free charges are on the metal plates and dielectric contains only dipoles. The net, unbalanced bound charge at the metal/dielectric interface balances the charge on the metal plate. If the dielectric is replaced by a doped semiconductor or an ionised gas, etc, then electrons move relative to the ions, and if the system is finite they both contribute to $\rho _{\text{f}}$ at the edges.^[1]^[2]

D is not determined exclusively by the free charge. AsE has a curl of zero in electrostatic situations, it follows that $\nabla \times \mathbf {D} =\nabla \times \mathbf {P}$

The effect of this equation can be seen in the case of an object with a "frozen in" polarization like a barelectret, the electric analogue to a bar magnet. There is no free charge in such a material, but the inherent polarization gives rise to an electric field, demonstrating that theD field is not determined entirely by the free charge. The electric field is determined by using the above relation along with other boundary conditions on thepolarization density to yield the bound charges, which will, in turn, yield the electric field.^[1]

In alinear,homogeneous,isotropic dielectric with instantaneous response to changes in the electric field,P depends linearly on the electric field, $\mathbf {P} =\varepsilon _{0}\chi \mathbf {E} ,$ where the constant of proportionality $\chi$ is called theelectric susceptibility of the material. Thus $\mathbf {D} =\varepsilon _{0}(1+\chi )\mathbf {E} =\varepsilon _{0}\varepsilon _{r}\mathbf {E} =\varepsilon \mathbf {E}$ whereε_r = 1 +χ is therelative permittivity of the material, andε is thepermittivity.

In linear, homogeneous, isotropic media,ε is a constant. However, in linearanisotropic media it is atensor, and in nonhomogeneous media it is a function of position inside the medium. It may also depend upon the electric field (nonlinear materials) and have a time dependent response. Explicit time dependence can arise if the materials are physically moving or changing in time (e.g. reflections off a moving interface give rise toDoppler shifts). A different form of time dependence can arise in atime-invariant medium, as there can be a time delay between the imposition of the electric field and the resulting polarization of the material. In this case,P is aconvolution of theimpulse response susceptibilityχ and the electric fieldE. Such a convolution takes on a simpler form in thefrequency domain: byFourier transforming the relationship and applying theconvolution theorem, one obtains the following relation for alinear time-invariant medium: $\mathbf {D} (\omega )=\varepsilon (\omega )\mathbf {E} (\omega ),$ where $\omega$ is the frequency of the applied field. The constraint ofcausality leads to theKramers–Kronig relations, which place limitations upon the form of the frequency dependence. The phenomenon of a frequency-dependent permittivity is an example ofmaterial dispersion. In fact, all physical materials have some material dispersion because they cannot respond instantaneously to applied fields, but for many problems (those concerned with a narrow enoughbandwidth) the frequency-dependence ofε can be neglected.

At a boundary, $(\mathbf {D_{1}} -\mathbf {D_{2}} )\cdot {\hat {\mathbf {n} }}=D_{1,\perp }-D_{2,\perp }=\sigma _{\text{f}}$ , whereσ_f is the free charge density and the unit normal $\mathbf {\hat {n}}$ points in the direction from medium 2 to medium 1.^[1]

History

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The earliest known use of the term is from the year 1864, in James Clerk Maxwell's paperA Dynamical Theory of the Electromagnetic Field. Maxwell introduced the termD, specific capacity of electric induction, in a form different from the modern and familiar notations.^[3]

It wasOliver Heaviside who reformulated the complicated Maxwell's equations to the modern form. It wasn't until 1884 that Heaviside, concurrently with Willard Gibbs and Heinrich Hertz, grouped the equations together into a distinct set. This group of four equations wasknown variously as the Hertz–Heaviside equations and the Maxwell–Hertz equations, and is sometimes still known as the Maxwell–Heaviside equations; hence, it was probably Heaviside who lentD the present significance it now has.

Example: Displacement field in a capacitor

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A parallel plate capacitor. Using an imaginary box, it is possible to use Gauss's law to explain the relationship between electric displacement and free charge.

Consider an infinite parallel platecapacitor where the space between the plates is empty or contains a neutral, insulating medium. In both cases, the free charges are only on the metal capacitor plates. Since the flux linesD end on free charges, and there are the same number of uniformly distributed charges of opposite sign on both plates, then the flux lines must all simply traverse the capacitor from one side to the other. InSI units, the charge density on the plates is proportional to the value of theD field between the plates. This follows directly fromGauss's law, by integrating over a small rectangular box straddling one plate of the capacitor:

$\oiint$

\scriptstyle _{A}

\mathbf {D} \cdot \mathrm {d} \mathbf {A} =Q_{\text{free}}

On the sides of the box, dA is perpendicular to the field, so the integral over this section is zero, as is the integral on the face that is outside the capacitor whereD is zero. The only surface that contributes to the integral is therefore the surface of the box inside the capacitor, and hence $|\mathbf {D} |A=|Q_{\text{free}}|,$ whereA is the surface area of the top face of the box and $Q_{\text{free}}/A=\rho _{\text{f}}$ is the free surface charge density on the positive plate. If the space between the capacitor plates is filled with a linear homogeneous isotropic dielectric with permittivity $\varepsilon =\varepsilon _{0}\varepsilon _{r}$ , then there is a polarization induced in the medium, $\mathbf {D} =\varepsilon _{0}\mathbf {E} +\mathbf {P} =\varepsilon \mathbf {E}$ and so the voltage difference between the plates is $V=|\mathbf {E} |d={\frac {|\mathbf {D} |d}{\varepsilon }}={\frac {|Q_{\text{free}}|d}{\varepsilon A}}$ whered is their separation.

Introducing the dielectric increasesε by a factor $\varepsilon _{r}$ and either the voltage difference between the plates will be smaller by this factor, or the charge must be higher. The partial cancellation of fields in the dielectric allows a larger amount of free charge to dwell on the two plates of the capacitor per unit of potential drop than would be possible if the plates were separated by vacuum.