Complex relative permittivity is defined as

$$\epsilon_r = \frac{\epsilon(\omega)}{\epsilon_0}=\epsilon_r^{\prime}(\omega) + i\epsilon_r^{\prime\prime}(\omega) = \epsilon_r^{\prime}(\omega) + \frac{i\sigma}{\omega\epsilon_0}$$

Measuring the real static permittivity, $\epsilon_r(\omega=0)$, can be done by a plate capacitor, $\epsilon_r=\frac{C_x}{C_0}$. But how can I get a chart like this one?

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EDIT: A practical question about the same thing was posted on electronics.SE.


First make a parallel plate capacitor with plates of area A and spacing d. Fill the space between the plates with the dielectric whose complex permittivity $\epsilon(\omega)$ you wish to measure.

The formula for this capacitance is a complex function of frequency because the permittivity is a complex function of frequency. $$ C(\omega)={{\epsilon(\omega)A}\over{d}} $$ This capacitor has a complex impedance $Z(\omega)$ which is also a function of frequency $$ Z(\omega)={1\over {i\omega C(\omega)}}={1 \over {i\omega {{\epsilon(\omega)A}\over d}}} $$ Rearranging gives $$ \epsilon(\omega)={d\over {i\omega Z(\omega)A}} $$

Now from a signal generator put a voltage $V sin(\omega t)$ across Z, measure the amplitude ($I(\omega)$) and phase shift ($\delta(\omega)$) of the current $I(\omega) sin(\omega t - \delta(\omega))$ that flows through Z, and use Ohm's law to calculate the complex impedance. $$ Z(\omega)={V \over {I(\omega)}}e^{i\delta (\omega)} $$ Plug this into the previous equation to get the complex $\epsilon(\omega)$.

There are commercial instruments called LCR Meters or Network Analyzers which have a signal generator built in and are used to measure complex impedances as described above.

This method will work to about 1 GHz. Studying the transmission and reflection in a dielectric filled wave guide will work for about 1-10 GHz. Infrared, visible, and ultraviolet light transmission and reflection from the dielectric can be used at higher frequencies. Remember the index of refraction is also complex and related to $\epsilon$ by: $$ n = \sqrt{\epsilon \mu} $$ The materials are often non-magnetic so the magnetic permeability $\mu$ is very close to the constant $\mu_0$.


Use the wave equation $(\frac{1}{\epsilon \epsilon_0 \mu \mu_0} \vec{\nabla}^2 - \frac{\partial^2}{\partial t^2}) A = 0$ and its solution, i. e. plane waves $A = A_0 \exp(\imath ( \omega t - \vec{k} \vec{x}))$ with $||\vec{k}|| = \frac{\omega}{\sqrt{\epsilon \epsilon_0 \mu \mu_0}}$. In words: The bit you already knew was that the (real part of the) wavector, $\mathrm{Re} (||\vec{k}||)$ (and hence the wavelength), indicates the real part $\epsilon^\prime$ of $\epsilon = \epsilon^\prime + \imath \epsilon^{\prime\prime}$. The bit you see here just as naturally is that its imaginary part is related to $\epsilon^{\prime\prime}$. It causes the otherwise purely oscillatory exponential to contain an exponential decay. By measuring its scale length (or the attenuation and calculate it from that), you can get the quantity you were after optically. In a frequency range or with absorption coefficients where you do not operate with waves, you can measure an analogous loss in whatever setup (plate capacitor?) you choose for that.


The imaginary part of the permittivity is a measure of the loss in the system. In some literature you will see reference to the "loss tangent" - this is in essence the ratio of the real and imaginary parts of the dielectric constant. The simplest way to measure this is to put the sample of interest in a cavity (for example, make it the dielectric of a capacitor) and sweep the frequency applied. Then you measure the amplitude and phase of the current.

For a perfect dielectric, the imaginary component of the dielectric constant is zero, and there will be a 90° phase difference between the voltage applied and the current flowing. However, the moment you have a lossy medium, the phase angle changes (towards zero degrees: a resistor is a "perfectly lossy capacitor" in this sense).

Phase and amplitude of a signal can be measured quite accurately. From this you get the complex impedance.

Relating this back to the properties of the dielectric is as simple as

$$Z = \frac{1}{j\omega C} = \frac{V}{I}$$ where

$$C = \frac{\epsilon A}{d}$$

Combining these, we get

$$\epsilon(\omega) = \frac{d}{j \omega Z A} = \frac{d\cdot I}{j\omega V A}$$

Where $I$ and $V$ are complex numbers. If you just know the phase angle $\phi$ between $I$ and $V$, you can simplify the math a little bit by writing as

$$\epsilon(\omega) = \frac{d\cdot |I|e^{j\phi}}{j\omega |V| A}$$

When $\phi=\frac{\pi}{2}$, all the complex terms disappear and you are left with a simple expression (real dielectric constant, i.e. no dissipation); as $\phi$ becomes less than that, the imaginary term appears.

  • $\begingroup$ Nit: In the first sentence I think "complex permittivity" should be "imaginary part of the permittivity". $\endgroup$ – DanielSank Oct 7 '15 at 23:35
  • $\begingroup$ Rolled back edit after incorrect and unnecessary addition. $\endgroup$ – Jold Jan 9 '16 at 19:53
  • $\begingroup$ @jld why incorrect and why unnecessary? $\endgroup$ – Sparkler Jan 9 '16 at 20:04
  • $\begingroup$ @Sparkler: $e^{i\phi}=\cos\phi+i\sin\phi$ is the correct relation, you had a minus cosine there. It's not really a necessary addition because it doesn't really add anything that wasn't known. $\endgroup$ – Kyle Kanos Jan 10 '16 at 12:29
  • 1
    $\begingroup$ @KyleKanos the expression in the equation is divided by j, that's why there's a minus. $\endgroup$ – Sparkler Jan 10 '16 at 14:46

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