The electric flux density is defined as $$\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$$ where P is the polarization vector of the material. As I understand it, the net electric field includes the polarization component, and we define D in such a way that it is independent of the material or the bound charge. But if D is truly independent of the bound charge, then why does it change across boundaries of different materials? In particular, the tangential component changes from one material to another: $$\frac{D_{t1}}{\epsilon_1} = \frac{D_{t2}}{\epsilon_2}$$

I am also confused as to why P is added in the equation. It seems to me that if D is to be independent of P, then it should be subtracted instead.


2 Answers 2


Gauss Law. Polarization Vector

The Gauss Law brings the local relation between the electric field and the sources.
The main sources of electric field are free charges, but we can also consider the contribution of the field produced by the polarized material.

$$ \nabla\cdot\mathbf{E}(\mathbf{r}) =\dfrac{\rho_l(\mathbf{r}) + \rho_P(\mathbf{r})}{\varepsilon_0} $$

where \$\rho_l\$ is the free charge contribution, and \$\rho_P\$ is the contribution due to polarization. But

$$ \rho_P(\mathbf{r})=-\nabla\cdot\mathbf{P}(\mathbf{r}) $$

where \$\mathbf{P}(\mathbf{r})\$ is the Polarization Vector. Then

$$ \nabla\cdot\mathbf{E}(\mathbf{r})=\dfrac{\rho_l(\mathbf{r})-\nabla\cdot\mathbf{P}(\mathbf{r})}{\varepsilon_0} $$

The free charge density is

$$ \rho_l(\mathbf{r})=\nabla\cdot\left(\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r})\right) $$

remember that \$\nabla\cdot\mathbf{D}(\mathbf{r})=\rho_l(\mathbf{r})\$, we can write the general form of the Gauss Law:

$$ \mathbf{D}(\mathbf{r})=\varepsilon_0\,\mathbf{E}(\mathbf{r})+\mathbf{P}(\mathbf{r}) $$

The displacement vector \$\mathbf{D}(\mathbf{r})\$ is the combination of the applied field \$\mathbf{E}\$ and induced field \$\mathbf{P}\$ in the material by the polarization of its molecules. The polarization of a material depends on the external field, and in turn creates an induced field which overlaps the external field. Then there is a relationship between these fields, in particular between the polarization vector and total field (the field that can be measured).

For linear dielectrics (which are the most technological interest materials) applies: \$\mathbf{P}(\mathbf{r})=\chi_e\,\varepsilon_0\,\mathbf{E}(\mathbf{r})\$ and then

$$ \mathbf{D}(\mathbf{r})=\varepsilon_0(1+\chi)\mathbf{E}(\mathbf{r})=\varepsilon_0\,\varepsilon_r\,\mathbf{E(r)}=\varepsilon\,\mathbf{E(r)} $$

where \$\chi\$ is the dielectric susceptibility of the material. \$\varepsilon=\varepsilon_0\,\varepsilon_r=\varepsilon_0(1+\chi)\$ is the permittivity of material, and \$\varepsilon_r\$ is the relative permittivity.

The greater the permittivity of the material, is polarized more strongly and electrical effects are greater.

  • "D is the field due to the free charge only"
  • "E is the field due to both the free and bound charge"

  • the effect of the bound charge is included indirectly through the relative permeability


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.