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This question already has an answer here:

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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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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.

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