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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.

## 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.