The relation between permittivity and conductivity

Question

I am able to measure relative permittivity $\varepsilon_r$ of a fluid, and I want to calculate conductivity of same fluid. Can anyone suggest me how to do this?

I found one formula to calculate conductivity but I am not sure it is right or wrong. The formula is $$\varepsilon=\varepsilon_r\varepsilon_0 + j\frac\sigma\omega,$$ but i know that $$\varepsilon=\varepsilon_r\varepsilon_0 .$$

If this is equal then $\sigma$ should be zero according to this formula. I think the above one is wrong.

Can anyone tell me how to calculate a material's conductivity from its permittivity?

Answer 1

In general, you do not have enough information to do this. A material's permittivity and its conductivity are fundamentally different quantities, and cannot generally be calculated from one another, particularly if all you have are static measurements.

The main reason for this is that a material's permittivity is a purely static property, while its conductivity involves its dynamics. (One way to see this is in their units: $[\sigma/\varepsilon]=1\textrm{ s}^{-1}$, so any connection between them must involve a specific timescale.) As Purcell makes quite clear, conductance is a time-dependent phenomenon, and the distinction between insulators and conductors depends on what timescale you're considering, much like distinguishing solids and liquids depends on the timescale.

That said, a material's permittivity and its conductivity can be very productively integrated into a single, complex quantity. (Confusingly, this is often called its permittivity, with little distinction.) It is important to note that this is only ever done at one specific frequency $\omega$. In this case we write $$\epsilon(\omega)=\epsilon'(\omega)+j\epsilon''(\omega)=\epsilon_r(\omega)\epsilon_0+j\frac{\sigma(\omega)}{\omega},\tag{1}$$ where $j^2=-1$. This is very useful for oscillating fields, even in the quasi-static regime, as it integrates and simplifies a good number of formulae, but it is not very useful for DC fields. (Why the complex numbers? Oscillating fields are often represented as $\mathbf E=\mathbf E_0 e^{j\omega t}$, which puts $j$ into the game. If $\omega=0$ then the game is over.)

It is important to stress that equation $(1)$ is the correct one. For DC fields or where $\sigma$ can be neglected, we usually shorten it to $\varepsilon=\varepsilon_r\varepsilon_0$, but the latter is only an approximation of the former.

Having said all of that, I must add that there is one way to calculate a material's conductivity from its permittivity. This relies on the fact that, because of causality considerations, $\varepsilon(\omega)$ must be an analytic function for complex $\omega$ in the upper half-plane. Because of that, its real and imaginary parts must obey what are known as Kramers-Kronig relations (see also here for more on why those relations hold), which let you calculate $\sigma(\omega)$ as $$ \sigma(\omega)= -\frac{\varepsilon_0\omega}{\pi} \mathcal P\int_{-\infty}^\infty \frac{\varepsilon_r(\omega')}{\omega'-\omega}\mathrm{d}\omega', $$ where $\mathcal P$ denotes the Cauchy principal value of the integral must be taken at the singularity at $\omega'=\omega$. This is, however, rarely a practical option, as you need to know the electrostatic response $\varepsilon_r(\omega)$ at all frequencies $\omega$, or be in a position where you know you can neglect all frequencies outside some interval. If all you have is a single, static measurement of $\varepsilon_r$, then this can't possibly work.

I hope this is enough to point you in the right direction (which is, of course, to go and measure the conductivity directly!).

Answer 2

The permittivity $ \epsilon(\omega)$ could be a complex quantity, and you can write it :

$\epsilon(\omega) = \epsilon'(\omega) + j\epsilon''(\omega)$, where j is the imaginary unit.

In the case of a material with conductivity, $\epsilon''(\omega)$ comes from conductivity : $\epsilon''(\omega)= \frac{\sigma(\omega)}{\omega}$

So, if you know the imaginary part of the permittivity, you know the conductivity.

Answer 3

I think if you can measure the permittivity, then you must have an LCR bridge. using the latter: 1- measure the impedance of your sample, if it is >100 $k\,\Omega$ then go to 3 2- if your impedance is <100 $k\,\Omega$ then measure the following parameters, using your LCR bridge, Cs, D or Q, or better theta. 3- measure Cp, D or Q, or better theta. Now you have some data for certain frequency range, you can start calculation.

4- if your D or Q values are within the limits of the bridge, it is easier, so I'll assume this for now & we'll call option A: using the cell of your measured sample,measure the same capacitance for the air, $C_s$ or $C_p$. Or you can calculate it using your sample cell dimensions.

Now * divide all the measured Cs or Cp by the value for the air. you can get the permittivity * multiply the value of D or 1/Q by all the permittivity values you have. you you have the Loss (Imaginary part of the permittivity). * Apply the relation conductivity=2PiF EoE''. then you're there. hope this is easy to follow. Kind regards

Answer 4

It helps to look at the problem in frequency and momentum space. The convention used here is $$ f(\omega, \vec{k}) = \int e^{i (\omega t - \vec{k} \cdot \vec{r})} f(t, \vec{r}) \, dt d^3\vec{r} $$

Considering the macroscopic Maxwell equations, the one that deals with current is the Ampere's law. $$ i (\vec{k} \times \vec{H} + \omega \vec{D}) = \vec{J}_f $$

There is also the Ohm's law that defines the conductivity. $$ \vec{J}_f = \sigma \vec{E} $$

Note that $\sigma$ is complex, can be a tensor, and is generally a function of $\omega$. E.g., in the Drude model, $$ \sigma(\omega) = \frac{\sigma_0}{1- i\omega\tau} $$

Assuming a linear material, how the material responds to an E field is described by the constituent relation. $$ \vec{D} = \epsilon \vec{E} $$

Therefore, Ampere's law can also be written as $$ i \left(\vec{k} \times \vec{H} + \omega \left(\epsilon + \frac{i}{\omega} \sigma \right) \vec{E}\right) = \vec{0} $$

Suppose you have solved some problems in a lossless $(\sigma = 0)$ medium, such as EM wave propagation. To quickly get the corresponding result in a lossy medium, we can effectively take $$ \epsilon \to \epsilon + \frac{i}{\omega} \sigma $$

Be aware that $\epsilon$ appears in Gauss's law too, where it is hidden in the displacement field, $$ i \vec{k} \cdot \vec{D} = \rho_f $$

so we need to check if this substitution produces any contradiction. Using the continuity equation $$ \vec{k} \cdot \vec{J}_f = \omega \rho_f $$

We get $$ i \vec{k} \cdot (\epsilon + \frac{i}{\omega} \sigma) \vec{E} = 0 $$

So all is good.