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I am having a real hard time understanding how I reach the expression that relates resistance and $\varepsilon''$ as the complex part of the dielectric constant.

I was trying to follow a clear and eloquent way of describing this. I understand how the idealized concept of a capacitor requires a $\pi/2$ phase shift between polarization and electric field which is related by the dielectric constant. If it is not $\pi/2$ , then expressing the the dielectric constant as a complex number will allow us to express that phase shift.

I know that not having a phase shift equal to $\pi/2$ means that I do not have a ideal capacitor, but also a related resistance.

My professors keep telling me that the following expression is a definition, not a deduction:

$$\varepsilon''=\sigma/{\varepsilon_0\omega} $$

i.e.

$$\varepsilon''=d/{A\varepsilon_0\omega R} $$ where d and A are thickness and contact area of the resistance R

But it seems to me that I should be able to explain it better... How?

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Complex conductivity is defined by the linear relation $$ \tilde{j} = \tilde{\sigma} \tilde{E} $$ between phasors $\tilde{j},\tilde{E}$ (current density and electric field), which is well obeyed in most materials for weak fields and low frequencies (below X-rays...).

Alternatively, one may describe conduction properties of the material by complex permittivity, which is defined by the relation $$ \tilde{ P} = (\tilde{\epsilon} - \epsilon_0)\tilde{E}, $$ which expresses the same linearity, since in oscillating fields, current density $j$ can be expressed as time derivative of the polarization potential $P$: $$ j = \partial_t P. $$ Both complex conductivity and permittivty are functions of frequency.

For phasors we have the same relation $$ \tilde{j} = \partial_t \tilde{P}; $$ since phasors oscillate as $e^{i\omega t}$, this leads to $$ \tilde{j} = i\omega \tilde{P}. $$ From this last relation and the definitions of $\tilde{\sigma},\tilde{\epsilon}$ follows the relation $$ \tilde{\sigma} = i\omega (\tilde{\epsilon} - \epsilon_0). $$ Mathematically, the two complex functions $\tilde{\epsilon},\tilde{\sigma}$ are equivalent. In practice, conductivity is more convenient for description of metals, since for static electric field $E$ they conduct steady current for which the relation $j = \sigma E$ remains valid. On the other hand, permittivity is better suited for description of dielectrics, since for static electric field these settle down in an electrically polarized state for which the relation $P = (\epsilon - \epsilon_0)E$ remains valid (if $P$ is understood as average electric moment of neutral element of unit volume).

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  • $\begingroup$ Excellent! This was exactly the type of logic that I was looking for. However, I now have a doubt: in the last relation I can express it in such as $\varepsilon_r = 1 + \sigma / i\omega \varepsilon_0$. It seems now that the real part of the dielectric constant $\varepsilon '$ is 1. Does this make sense? $\endgroup$ – cinico Feb 4 '14 at 23:28
  • $\begingroup$ In general, $\sigma$ is complex so the real part of $\epsilon_r$ depends on the imaginary part of $\sigma$. If by accident $\sigma$ is real, real part of $\epsilon_r$ is indeed 1. $\endgroup$ – Ján Lalinský Feb 4 '14 at 23:52
  • $\begingroup$ That's it :) Thank you so much!! I've been banging my head over the past days because of this $\endgroup$ – cinico Feb 5 '14 at 0:01
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I do not understand the situation. On the one hand you are talking about a dielectric (which is non-conducting by definition and hence has infinite resistance) and on the other hand you mention a conductivity $\sigma$ and resistance R. Now, in the general linear case, an absorptive medium is characterised by a complex, frequency-dependent $\epsilon(\omega)={\epsilon}'(\omega)+i{\epsilon}''(\omega)$, where ${\epsilon}''$ is responsible for losses (it is a sum of $\delta$-functions in a dispersive, non-absorptive medium). The polarisation current is given by $$j(t)=∂_{t}P(t)=∫_{t₀}^{t}dsχ′(t-s)E(s),χ(t)=0,t<0,$$ where $χ(t)$ is the permittivity. After Fourier transforming $$j(ω)=iωχ(ω)E(ω),χ(ω)=ɛ(ω)-ɛ_0.$$ I need more information to proceed. What is this $\sigma$, is it real or complex, does it depend on $\omega$? And what is R?

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  • $\begingroup$ I was talking about a general material, with both capacitance and resistance behaviour.I was told that the definition of $\varepsilon ''$ was dependent on the conductivity $\sigma$ (DC?). From the conductivity I can (having an Area and thickness of a sample) reach a Resistance. My problem is because the definition of $\varepsilon ''$ seems to have fallen from nowhere... $\endgroup$ – cinico Feb 4 '14 at 22:13
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The question raised is related to imperfect dielectic materials. It is a long-standing debat about the so-called Debye dielectrics (of the kind mention in some part of the preceeding answers) and even more complicated to understand, the non-Debye dielectrics (a better model for the real dielectrics).

There are a lot of publications on the subject and on a related special element (CPE) involved in the models : http://en.wikipedia.org/wiki/Constant_phase_element

Many perturbative phenomenas have been ascribed in attempt to explain the imperfect behaviour of real dielectrics. I made a compendium of some of them in the paper "A number of models for CPA of conductors and for relaxation in non-Debye dielectrics", Journal of Non Crystalline Solids, 121-133 (1991), Elsevier Sience Publischers. A updated version is available on the web : https://fr.scribd.com/doc/71923015/The-Phasance-Concept , pp.14-18.

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