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Electric polarization

  1. May 11, 2013 #1
    Does the electric polarization density of a dielectric inside a capacitor have the same direction as the electrical field? Considering the electric dipole moment vector goes from the - charge to + charge?
     
  2. jcsd
  3. May 11, 2013 #2

    vanhees71

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    Yes, because the electrons in the dielectric are dragged a bit out of their equilibrium positions due to the applied electric field in the opposite direction than the field, because the force is [itex]\vec{F}=-e \vec{E}[/itex], where [itex]-e<0[/itex] is the electron's charge. The net effect (in lineare-response approximation) is an induced polarization
    [tex]\vec{P}=\chi_{\text{el}} \vec{E}.[/tex] This is for homogeneous isotrophic media and time-independent (static) electric fields.

    For time-dependent fields, this relation holds in the frequency domain, i.e., you have
    [tex]\vec{P}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} t' \chi_{\text{el}}(t-t') \vec{E}(t',\vec{x}).[/tex]
    The causality constraint, [itex]\chi_{\text{el}}(t-t') \propto \Theta(t-t')[/itex] makes the susceptibility to a retarded Green's function. In the frequency domain, i.e., for the Fourier transform of the quantities the above convolution integral translates into
    [tex]\tilde{\vec{P}}(\omega,\vec{x})=\tilde{\chi}_{\text{el}}(\omega) \tilde{\vec{E}}(\omega,\vec{x}).[/tex]
    The retardation condition makes [itex]\tilde{\chi}(\omega)[/itex] a holomorphic function in the upper complex [itex]\omega[/itex]-half plane, where use the usual physicist's convention for Fourier transforms between the time and frequency domain:
    [tex]f(t)=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{2\pi} \tilde{f}(\omega) \exp(-\mathrm{i} \omega t).[/tex]
     
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