Electric displacement

Electric displacement $\mathbf{D}$ is a vector field that accounts for the electromagnetic effects of both the electric field generated by dielectric polarization and an external electric field. In other words, it handles both the source and response fields at the same time. It is defined as

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

where $\varepsilon_{0}$ is the vacuum permittivity, $\mathbf{E}$ is the external field and $\mathbf{P}$ is the polarization. Like polarization, it has the units of a surface charge distribution.

Unlike $\mathbf{E}$ , there is no "Coulomb's law" for displacement as the Curl of $\mathbf{D}$ is not necessarily zero:

\nabla\times\mathbf{D}=\nabla\times\mathbf{P}

It is only zero if the polarization is irrotational, and it usually isn't. An example of irrotational polarization is when the external charge density is zero, that is, there is no external field.

Gauss' law#

It obeys a form of Gauss' law that is similar to that of the electric field

\nabla\cdot\mathbf{D}=\rho_{f},\qquad \oint \mathbf{D}\cdot d\mathbf{a}=Q_{f,enc}

in both differential and integral forms. $\rho_{f}$ is the free charge density (the one responsible for the external field) and $Q_{f,enc}$ is the total electric charge of $\rho_{f}$ enclosed within the Gaussian surface.

Boundary conditions#

As the displacement is dependent on the electric field, it inherits the discontinuity over surface charges. From the integral form of Gauss' law we get

D_\text{above}^{\perp}-D_\text{below}^{\perp}=\sigma_{f}

and from the curl of $D$ we get

\mathbf{D}_\text{above}^{\parallel}-\mathbf{D}_\text{below}^{\parallel}=\mathbf{P}_\text{above}^{\parallel}-\mathbf{P}_\text{below}^{\parallel}

When dealing with dielectrics, these may be more meaningful than the boundary conditions on the electric field itself.