# What is the relation between permittivity and susceptibility of a dielectric? [duplicate]

This question might be trivial but I have a conceptual confusion regarding permittivity and susceptibility of a dielectric

According to Wikipedia, permittivity is the ability to resist external electric field. This means a substance with high permittivity requires high external electric field to polarize. On the other hand, susceptibility is defined as the ability to polarize. So a substance with higher susceptibility should polarize easily. This means both the quantities are inversely proportional to each other but mathematically they are linearly dependant

Where am I going wrong?

I get it that there is a similar question but the answer for that question did not answer mine and I still did not get the privilege of commenting as I am a newbie. That's why I posted it as a separate question

I understood the definitions but couldn't understand the mathematical relation

## marked as duplicate by sammy gerbil, Jon Custer, stafusa, Mitchell, John RennieNov 11 '17 at 8:22

permittivity can be thought of as a the dielectric constant for a material between the two plates of a capacitor. You can find a simple capacitor and work out an impedance for your choosen circuit. The inverse of impedance is something called the admittance. The imaginary part of the admittance is susceptance.

As the comment says, it's answered there.

But, since it's true there's a lot of confusion (because it is usually too messingly-explained), let's give a try. The key is: add formulas. Maths are thought to be confusing and they make you lose perspective, but I instead find them to show where things come from in a comapct way.

The vector D is defined to be $\vec{D}=\varepsilon_0\vec{E}+\vec{P}$, where P is the polarisation.

Now, we make a first assumption (basic case): polarisation is linear with the electric field $\vec{P}=\alpha \vec{E}$ (it doesn't really have to be linear, but let's make it easier, just to show). Let's call $\chi_e=\alpha/\varepsilon_0$ so that $\alpha=\chi_e\cdot\varepsilon_0$ and $P=\varepsilon_0\chi_e E$

Now, taking common factor,

$D=\varepsilon_0(1+\chi_e)E$

We call $\varepsilon:= \varepsilon_0(1+\chi_e)$.

So, the permitivity is $\varepsilon$, and it is the relation bteween the electric field and the displacement vector. Susceptibility is $\chi_e$ and it is strongly related to it. It's, roughly speaking, like "the permitivity of the polarisation part only$. So susceptibility is related to polarisation, and the permitivity just connects$\vec{D}$and$\vec{E}$. Note: in the general case,$P$doesn't have to be linear with the electric field. In that case, we "introduce the non-linearities" in the$\chi_e\$ expression. Moreover, it can be a function of the point considered, and even a matrix or combination of tensors.