Electric flux density

The electric flux density $\mathbf{D}$ is the vector field

$\mathbf{D}=ϵ\mathbf{E}{\text{(C/m}}^2\text{)},$

where $\mathbf{E}$ is the Electric field and $ϵ={ϵ}_r{ϵ}_0$ is the Permittivity of the medium. Its units are coulombs per square meter — explicitly a charge per area, the same as a surface charge density.

The point of $\mathbf{D}$ is that its sources are exactly the free charges. The defining differential relation is Gauss’s law:

$\nabla \cdot \mathbf{D}={\rho }_v,$

where ${\rho }_v$ is the free volume charge density. Note that ${\rho }_v$ on the right is only free charge — bound polarization charges in dielectrics are absorbed into $ϵ$ via $\mathbf{D}=ϵ\mathbf{E}$. By contrast, $\nabla \cdot \mathbf{E}={\rho }_{\text{total}}/{ϵ}_0$ involves all charge (free + bound), which is harder to compute when dielectrics are present.

Geometric meaning

The integral form of Gauss’s law,

${\oint }_S\mathbf{D}\cdot d\mathbf{s}=Q_{\text{enc}},$

says: the flux of $\mathbf{D}$ through any closed surface equals the total free charge enclosed. So $\mathbf{D}$ literally has “flux lines” that begin on positive free charges and end on negative free charges — they don’t terminate on bound polarization charges within dielectrics. That’s why $\mathbf{D}$ is sometimes called the electric displacement field.

The unit (C/m²) also makes intuitive sense: spread $Q_{\text{enc}}$ over the enclosing surface, and you get a flux density.

Why two fields?

$\mathbf{E}$ is the “what a test charge feels” field. $\mathbf{D}$ is the “what the free sources put out” field. They differ by $ϵ$:

• Inside a dielectric, the free sources push out a certain $\mathbf{D}$. Bound dipoles in the dielectric partially cancel the field, so a test charge actually feels $\mathbf{E}=\mathbf{D}/ϵ$, which is smaller than $\mathbf{D}/{ϵ}_0$ would suggest.
• At boundaries between two dielectrics, $\mathbf{E}$ has a discontinuous normal component (because ${\mathbf{D}}_n$ is continuous if there’s no free surface charge, so ${\mathbf{E}}_n=D_n/ϵ$ jumps when $ϵ$ changes).

See Electric boundary conditions for the full set of jump rules.

Relation to polarization

The microscopic picture, sometimes presented before $\mathbf{D}$ is introduced:

$\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P},$

where $\mathbf{P}$ is the polarization — dipole moment per unit volume in the dielectric. For a linear medium, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and so $\mathbf{D}={ϵ}_0(1+{\chi }_e)\mathbf{E}=ϵ\mathbf{E}$, recovering the simple constitutive relation.

In Maxwell’s equations

$\mathbf{D}$ appears in two of Maxwell’s four equations:

• Gauss’s law: $\nabla \cdot \mathbf{D}={\rho }_v$.
• Ampère-Maxwell: $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$. The ${\partial }_t\mathbf{D}$ term is the Displacement current — the conceptual breakthrough that made electromagnetic waves predictable.

The pair $\mathbf{D},\mathbf{H}$ are the “free-source” fields; the pair $\mathbf{E},\mathbf{B}$ are the “total-effect” fields. Constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$ link them through the material parameters.