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What is free charge

  1. Jul 24, 2014 #1

    Charge and current are either free or bound.

    Bound charge is charge which is displaced locally (that is, slightly), and bound current is current which loops locally (such as an electron "orbiting" a nucleus).

    In fluid mechanics, free charge and current are analogous to flow, while bound charge is analogous to pressure and viscosity, and bound current is analogous to vorticity.

    In solid mechanics, an external force may move material as a whole, analogously to free charge, and may also cause stress within material, analogously to bound charge.

    There are three versions of [itex]Gauss'\ Law[/itex] and the [itex]Ampere-Maxwell\ Law[/itex] (two of Maxwell's equations): the [itex]free[/itex] version, the [itex]bound[/itex] version, and the [itex]total[/itex] version.

    Each version has its own pair of electric and magnetic fields, and its own current and charge.


    [tex]\varepsilon_0\,\mathbf{E}\ =\ \frac{1}{\mu_0\,c^2}\,\mathbf{E}\ =\ \mathbf{D}\ -\ \mathbf{P}[/tex]

    [tex]\frac{1}{\mu_0}\,\mathbf{B}\ =\ \mathbf{H}\,+\ \,\mathbf{M}[/tex]

    Note that since the bound electric field is (usually) in exactly the [itex]opposite[/itex] direction to the free and total fields [itex]\mathbf{D}\text{ and }\mathbf{E}[/itex] (since it is caused by the molecular dipoles [itex]opposing\ \mathbf{D}[/itex]), it is conventionally written as [itex]minus\ \mathbf{P}[/itex], so that all three letters represent fields in (usually) the same direction.

    [tex]\mathbf{J}\ =\ \mathbf{J}_f\ +\ \mathbf{J}_b[/tex]

    [tex]\rho\ =\ \rho_f\ +\ \rho_b[/tex]

    [tex]\mathbf{P}\ = \chi_e\,\varepsilon_0\,\mathbf{E}[/tex]

    [tex]\mathbf{M}\ = \chi_m\,\mathbf{H}\ = \frac{1}{\mu_0}\,\chi_m\,(\chi_m\,+\,1)^{-1}\,\mathbf{B}[/tex]

    Extended explanation

    Gauss' Law and the Ampére-Maxwell Law may be expressed in three versions:

    The free version involves fields [itex]\mathbf{D}[/itex] and [itex]\mathbf{H}[/itex], free current density [itex]\mathbf{J}_f[/itex] and free charge density [itex]\rho_f[/itex].

    The bound version involves fields [itex]-\mathbf{P}[/itex] and [itex]\mathbf{M}[/itex], bound charge density [itex]\mathbf{J}_b[/itex] and bound current density [itex]\rho_b[/itex].

    The total version involves fields [itex]\mathbf{E}[/itex] and [itex]\mathbf{B}[/itex], total current density [itex]\mathbf{J}[/itex] and total charge density [itex]\rho[/itex].

    Expressed in differential forms, with [itex]\text{d}\ =\ \left(\frac{1}{c}\frac{\partial}{\partial t}\,,\,\nabla\right)[/itex], they are:

    [tex]\text{d}\ (c\mathbf{D}; \mathbf{H})^*\,=\,(c\,\nabla \cdot \mathbf{D}\ ,\nabla\times \mathbf{H}\,+\,c\,\frac{ \partial\mathbf{D}}{ \partial t})^*\,=\,(c\rho_f , \mathbf{J}_f)^*[/tex]

    [tex]\text{d}\ (-c\mathbf{P};\mathbf{M})^*\,=\,(-c\,\nabla \cdot \mathbf{P}\ ,\nabla\times\mathbf{M}\,-\,c\,\frac{\partial\mathbf{P}}{\partial t})^*\,=\,(c\rho_b , \mathbf{J}_b)^*[/tex]​

    and their sum:

    [tex]\text{d}\ (\varepsilon_0 c\mathbf{E};\frac{1}{\mu_0}\mathbf{B})^*\,=\, \frac{1}{\mu_0}\,\text{d}\ (\frac{1}{c}\mathbf{E};\mathbf{B})^*\,=\, \frac{1}{\mu_0}\,(\frac{1}{c}\nabla \cdot \mathbf{E}\ ,\ \nabla\times\mathbf{B}\,-\ \frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t})^*\ =\ (c\rho , \mathbf{J})^*[/tex]​

    By comparison, Gauss' Law for Magnetism and Faraday's Law have only one version:

    [tex]\text{d}\ (\frac{1}{c}\mathbf{E};\mathbf{B})\ =\ (\nabla \cdot \mathbf{B}\ ,\ \frac{\partial\mathbf{B}}{\partial t}\,+\,\nabla\times\mathbf{E})^*\,=\,0[/tex]

    The direction of the bound electric field, P:

    In isotropic materials, the total and free electric fields [itex]\mathbf{D}\text{ and }\mathbf{E}[/itex] are always in the same direction, and the bound field is always in the opposite direction.

    This is because the bound field is caused by the molecular dipoles opposing the free field.

    So the free field is always larger (in magnitude) than the total field: [itex]|\mathbf{D}| = \varepsilon_0|\mathbf{E}| + |\mathbf{P}|[/itex], and so it is more convenient to write the bound field as minus [itex]\mathbf{P}[/itex], so that [itex]\mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P}[/itex]. :wink:


    Permittivity (the relation between D and E) is generally a tensor, not a scalar, and so the D and E fields are generally not in exactly the same direction.

    However, most materials are isotropic, meaning that the permittivity tensor is a multiple of the unit tensor (in other words: the permittivity is a scalar), and so the D and E fields are in the same direction.

    Even in non-isotropic materials, however, the bound electric field still opposes the free field, and is still approximately opposite to it, and the free field is still larger than the total field.


    Since the free-field [itex]\mathbf{D}[/itex] is obtained by removing the bound-field [itex]-\mathbf{P}[/itex] from the total field [itex]\varepsilon_0\mathbf{E}[/itex], it follows that [itex]\mathbf{P}[/itex] deals with local displacement, while [itex]\mathbf{D}[/itex] deals with everything else.

    The name "electric displacement field" for [itex]\mathbf{D}[/itex] is therefore misleading, and would be better applied to [itex]\mathbf{P}[/itex].

    A better name for [itex]\mathbf{D}[/itex] would be the "electric free-field". Similarly, the magnetic intensity [itex]\mathbf{H}[/itex] might better be called the "magnetic free-field".

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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