Maxwell equations are the fundamentals of electrodynamics[1].

Maxwell’s Equations in Vacuum

Differential Form

$\left. \begin{array}{l l} \nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0} & Gauss's\ Law\\ \nabla\cdot\vec{B}=0 & Gauss's\ Law\ for\ Magnetism\\ \nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t} & Faraday's\ Law\\ \nabla\times\vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t} & Ampere's\ Law\ with\ Maxwell's\ Addition \end{array} \right\} I$

Notice, Gauss’s law for magnetism means that for magnetic field there is no magnetic charge like the case of electric charge. Which also means that for the magnetic field there is no source or drain. Where $\epsilon_0$ and $\mu_0$ are the permittivity and permeability of free space.

Integral Form

$\begin{array}{l} \iint\vec{E}\cdot d\vec{s}=\frac{1}{\epsilon_0}\iiint\rho d\tau\\ \iint\vec{B}\cdot d\vec{s}=0\\ \oint\vec{E}\cdot d\vec{l}=-\frac{d\Psi}{dt}\\ \oint\vec{B}\cdot d\vec{l}=\mu_0 I+\mu_0 \epsilon_0\iint\frac{\partial\vec{E}}{\partial t}\cdot\ d\vec{s} \end{array}$

Maxwell’s Equations in Dielectric Media

Electric polarization

For the magnetic fields in a dielectric media, there exist additional effect of electric polarization. Where the polarization vector $\vec{P} = \epsilon_0 \chi_e \vec{E}$. And the displacement field vector is defined as: $\vec{D} = \vec{P} + \epsilon_0\vec{E} = \epsilon_0(1 + \chi_e)\vec{E} = \epsilon_0 \epsilon_r \vec{E} = \epsilon\vec{E}$, where: $\epsilon$ is the absolute permittivity of the media , $\epsilon_r = 1 + \chi_e$ is the relative permittivity of the media, $\chi_e$ is the polarization rate of the media.

Magnetic polarization

For a media, with the presence of Magnetic field, similar to electric polarization, it will also suffer from a magnetic polarization. This means the magnetic field in the media will not be the original applied field strength. So we define the magnetizing field as: $\vec{H} = \frac{1}{\mu}\vec{B} = \frac{1}{\mu_0\mu_r}\vec{B}$

Differential Form

$\left. \begin{array}{l l} \nabla\cdot\vec{D}=\rho & Gauss's\ Law\\ \nabla\cdot\vec{B}=0 & Gauss's\ Law\ for\ Magnetism\\ \nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t} & Faraday's\ Law\\ \nabla\times\vec{H} = \vec{J} + \frac{\partial\vec{D}}{\partial t} & Ampere's\ Law\ with\ Maxwell's\ Addition \end{array} \right\} IV$

General case of Maxwell equations

$\left. \begin{array}{l l} \nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0} & Gauss's\ Law\\ \nabla\cdot\vec{B}=0 & Gauss's\ Law\ for\ Magnetism\\ \nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t} & Faraday's\ Law\\ \nabla\times\vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t} & Ampere's\ Law\ with\ Maxwell's\ Addition \end{array} \right\} V$

For equations V, if we consider $\rho$ and $\vec{j}$ as the total charge density and current density (including the polarization effect) with $\rho_e = \rho_{e_f} + \rho_{e_P}$ and $\vec{j} = \vec{j_f} + \vec{j_P} + \vec{j_M}$ (where $f, P, M$ each represent free, polarization and magneticsim). Then equations V is equivalent to equations IV for Maxwell’s equations in all medias. For example, plasma is a good conductor of electricity, thus $\rho_P = 0$, $\vec{j_P} = 0$ but for dielectric media $\vec{j_P} \neq 0$.

Why need displacement field vector $\vec{D}$

$\vec{D}$ is the introduced by Maxwell to the original Ampere’s law. But why should Maxwell introduce such a physical value to this equation [2]. First, we consider the Ampere’s law in the static condition $\oint\vec{H}\cdot d\vec{l} = I_0$. It is satisfied in all medias, but it fails under the case of changing field. When the electric field is changing in a system, eg: an electric circuit with an electric charger. We introduce two surfaces in the circuit where the two surfaces share the same Ampere loop $L$. $S_1$ is placed across one cross section of the wire $S_2$ is placed between the plates of the charger, Thus apply the original Ampere’s law, during the charing process of this circuit, we have: $\oint\vec{H}\cdot\vec{l} = I_0 = \iint_{S_1}\vec{j_1}\cdot d\vec{s}\neq 0 = \iint_{S_2}\vec{j_2}\cdot d\vec{s}$. Because in the wire current density $\vec{j_1}\neq 0$ but between the plates of the charger current density $\vec{j_2} = 0$. Thus Ampere’s law is no longer satisfied for the current changing circuit. To solve this problem, we look into the past of the Gauss’s law for charged particles $\iint\vec{D}\cdot d\vec{s} = q_0$. And considering the fact that current $I = \frac{dq}{dt}$, then we have $I_0 = \frac{d}{dt}\iint\vec{D}\cdot\vec{s}$. If we put this part into the equation of Ampere’s law to makeup for no wire part of current. We have the Maxwell’s modified Ampere’s law as: $\oint\vec{H}\cdot d\vec{l} = I_0 + \frac{d}{dt}\iint\vec{D}\cdot\vec{s}$. Convert this term to the derivative form with Stocks’s law we have: $\iint\nabla\times\vec{H}\cdot d\vec{s} = \iint\vec{j}\cdot d\vec{s} + \frac{d}{dt}\iint\vec{D}\cdot d\vec{s} = \Rightarrow \iint(\nabla\times\vec{H} - \vec{j} - \frac{\partial}{\partial t}\vec{D})\cdot d\vec{s} = 0$. To let this term hold constant we must have $\nabla\times\vec{H} = \vec{j} + \frac{d}{dt}\vec{D}$. Thus the derivative of Maxwell modified Ampere’s law is satisfied for both static and time changing conditions.

Reference

[1] https://en.wikipedia.org/wiki/Maxwell%27s_equations

[2] Kaihua Zhao, Ximo Chen, New Conception Physics Series: Electromagnetism (in Chinese), Higher Educational Press, Beijing, 2003 (赵凯华，陈熙谋，新概念物理学教程：电磁学，高等教育出版社，北京，2003)