Understanding Electric Polarization in Dielectric Capacitors

[Nikitin]

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?

 

[vanhees71]

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 $\vec{F}=-e \vec{E}$, where $-e<0$ is the electron's charge. The net effect (in lineare-response approximation) is an induced polarization
$$\vec{P}=\chi_{\text{el}} \vec{E}.$$ 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
$$\vec{P}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} t' \chi_{\text{el}}(t-t') \vec{E}(t',\vec{x}).$$
The causality constraint, $\chi_{\text{el}}(t-t') \propto \Theta(t-t')$ 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
$$\tilde{\vec{P}}(\omega,\vec{x})=\tilde{\chi}_{\text{el}}(\omega) \tilde{\vec{E}}(\omega,\vec{x}).$$
The retardation condition makes $\tilde{\chi}(\omega)$ a holomorphic function in the upper complex $\omega$-half plane, where use the usual physicist's convention for Fourier transforms between the time and frequency domain:
$$f(t)=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{2\pi} \tilde{f}(\omega) \exp(-\mathrm{i} \omega t).$$