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Electric displacement field



In physics, the electric displacement field or electric induction[citation needed] is a vector field \mathbf{D} that appears in Maxwell's equations. It accounts for the effects of unbound charges within materials. "D" stands for "displacement," as in the related concept of displacement current in dielectrics.

Additional recommended knowledge

Definition

In general, D is defined by the relation

\mathbf{D} = \varepsilon_{0} \mathbf{E} + \mathbf{P}

where E is the electric field, \varepsilon_{0} is the vacuum permittivity, and P is the polarization density of the material.

If P can be written as a linear function of E, which is the case in most materials, it can be written as

\mathbf{P} = \chi \varepsilon_{0} \mathbf{E},

and by

\mathbf{D} = \varepsilon_{0} \mathbf{E} + \mathbf{P} = \varepsilon_{0}(1 + \chi) \mathbf{E} \equiv \varepsilon \mathbf{E}

it yields

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

where \varepsilon is the permittivity of the material. In linear isotropic media this will be a constant, and in linear anisotropic media it will be a rank 2 tensor (a matrix). χ is called the electric susceptibility.

Displacement field in a capacitor

Consider an infinite parallel plate capacitor placed in space (or in a medium) with no free charges present except on the capacitor. In SI units, the charge density on the plates is equal to the value of the D field between the plates. This follows directly from Gauss's law, by integrating over a small rectangular box straddling both plates of the capacitor:

\oint_A \mathbf{D} \cdot d\mathbf{A} = Q

On the sides of the box, d\mathbf{A} is perpendicular to the field, so that part of the integral is zero, leaving, for the space inside the capacitor where the effect of the two plates adds:

|\mathbf{D}| = \frac{Q}{A}

which is the charge density on the plate. Outside the capacitor, the two plates effect compensate each other and |\mathbf{D}| = 0.

Units

In the standard SI system of units D is measured in coulombs per square meter (C/m²).

This choice of units results in one of the simplest forms of the Ampère-Maxwell equation:

\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}

If one chooses both B and H to be measured in teslas, and E and D to be measured in newtons per coulomb, then the formula is modified to be:

\nabla \times \mathbf{H} = \mu_0 \mathbf{J} + \frac{1}{c^2} \frac{\partial \mathbf{D}}{\partial t}

Therefore it is seen as being preferential to express B & H, and E & D in different sets of units.

Choice of units has differed in history, for instance in the electromagnetic system of scientific units, in which the unit of charge is defined so that 1 / 4\pi\varepsilon_0 = 1 (dimensionless), E and D are expressed in the same units.

 
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Electric_displacement_field". A list of authors is available in Wikipedia.
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