I have a question on the intuition behing the relation between relative permittivity $\epsilon(\omega)$ and electric susceptibility, $\chi_e$ of a of a material, more precise a dielectric medium.

First of all let fix a dielectric medium.

The relative permittivity $\epsilon(\omega)$ is given by $\varepsilon _{r}(\omega )={\frac {\varepsilon (\omega )}{\varepsilon _{0}}}$ where $ε(ω)$ is the complex frequency-dependent permittivity of the material, and $ε_0$ is the vacuum permittivity.

What is relative permittivity intuitively:

it measures the "answer" of dielectric medium to external electric field $E_{ext}$. That is a kind of 'sensitivity' for oncoming external sources.

The electric susceptibility $\chi_e$ intuitively indicates the degree of polarization of a dielectric material in response to an applied external electric field.

Mathematically relative permittivity & electric susceptibility are related via $$\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$$.

Question: Has this relation an intuitive meaning? I have only a kind of intuition for relative permittivity and electric susceptibility separately which I explaned above. But I don't find an heuristic explanation which justify $\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$. Can anybody expain the intuition one should have behind this relation?

Remark: I already have found this promising former discussion on similar problem What's the nuance of susceptibility and permittivity?

Why I think that it not provide the desired answer for my question? Well, the answers there are also treating the intuition behind permittivity and electric susceptibility separately. What I'm here looking for is the concrete intuition behind $\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$, ie how the are related. For example why $-1$ and not for example $-2$ or something like that.

Or if one see $\chi _{\text{e}}$ and $\varepsilon _{\text{r}}$ as kind of opposites, then one could intuitively think $\chi _{\text{e}}=-\varepsilon _{\text{r}}$ or $\chi _{\text{e}}=1-\varepsilon _{\text{r}}$ if one has transmission & reflection from optic in mind. By why $\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$?

  • $\begingroup$ "The electric susceptibility intuitively indicates the degree of polarization of a dielectric material in response to an applied external electric field." Actually, the electric susceptibility times $\epsilon_0$ does that. $\endgroup$
    – JEB
    Commented Jan 15, 2020 at 2:52
  • $\begingroup$ @JEB:so intuitively electric susceptibility is the capabitily to "absorb" the effect of the applied external electric field, that is the "affinity to be polarized", right? ok, that's the intuition for $\chi_e$. but that makes not fully clear the intuition behind $\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$. $\endgroup$
    – user267839
    Commented Jan 15, 2020 at 3:09
  • $\begingroup$ So in certain way we think $\chi _{\text{e}}$ and $\varepsilon _{\text{r}}$ as opposite to each other (compare with first two paragraphs of descheleschilder's answer in physics.stackexchange.com/questions/340626/…. But what is the meaning of $-1$ in $\chi _{\text{e}}\ =\varepsilon _{\text{r}}-1$? $\endgroup$
    – user267839
    Commented Jan 15, 2020 at 3:14

3 Answers 3


so my intuition is as follows:

relative permittivity: $$\epsilon_r=\frac{\epsilon}{\epsilon_0}$$ is the ratio between the medium permittivity and free space.

electric susceptibility: $$\chi_e=\epsilon_r-1=\frac{\epsilon-\epsilon_0}{\epsilon_0}$$ is the normalized difference between the medium permittivity and free space. "normalized" in this case being with respect to free space permittivity.

this might read as tautological but it really helps me separate them. the point is they're different ways of describing the same thing, one is a ratio representation, the other is a (normalized) delta.

i think any other explanation would boil down to a fancier version of the same explanation, ratio vs difference.


Refer to Purcell, Electricity and Magnetism, Chapter 10. The derivation is given with the best explanation.

This is a very important question and requires some clear thinking. This question is about year old and I am not sure the OP is still looking for an answer.

Imagine a setup where a battery $V$ is connected to a parallel plate capacitor, establishing a field $E$.

With vacuum the charge is $Q_0$, and with the dielectric between the plates the charge is $k$ times $Q_0$, which makes it $kQ_0$.

The dielectric will polarize itself to produce a charge $Q'$, to oppose the field.

This is the most important thing in the derivation - The field must stay the same $E$ no matter what happens between the metal plates, since the battery is $V$.

The $kQ_0$ on the plates adds to $(1-k)Q_0$ to make it appear as $Q_0$ between the plates for the battery.

$$Q_0 = kQ_0 + (1-k)Q_0$$

Now think of superposition for the fields.

The field $E''$ due $kQ_0$ adds to $E'$ due to $(1-k)Q_0$ to give the $E$ that should be the same field no matter what happens inside the plates.

$$E = E" + E'$$

$$E = kE - \frac {P}{\epsilon_0}$$

from this you can see

$$\frac{P}{\epsilon_0 E} = k-1$$

if you call susceptibility as the ratio $\frac{P}{\epsilon_0 E}$, then

$$\chi = \frac{P}{\epsilon_0 E} = k-1$$

$$\chi = k-1$$

$k$ is $\epsilon_r$


Why the field of the dielectric is $\frac{P}{\epsilon_0}$ ? for this you must read Purcell.


Macroscopic Maxwell equations require defining, in addition to fields $\mathbf{E},\mathbf{B}$, auxiliary fields $\mathbf{D},\mathbf{H}$, related via $$ \mathbf{D}=\epsilon_0\mathbf{E}+\mathbf{P},\\ \mathbf{H}=\frac{1}{\mu_0}\mathbf{B}-\mathbf{M}, $$ where $\mathbf{P},\mathbf{M}$ are respectively the polarization and the magnetization induced in the media.

Within the linear response framework/regime, the auxhiliary fields fields are related to the true fields via $$ \mathbf{D}=\epsilon\mathbf{E},\\ \mathbf{H}=\frac{1}{\mu}\mathbf{B}, $$ whereas the polarization and the magnetization are expressed as $$ \mathbf{P}=\epsilon_0\chi_e\mathbf{E},\\ \mathbf{M}=\chi_m\mathbf{H}, $$ The relations between permittivities and susceptibilities then easily follow (see also Electric susceptibility and Magnetic susceptibility where these derivations are given explicitly).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.