Questions about the relationship between capacitance and permittivity I am trying to understand the definition of capacitance, specifically, the relationship between permittivity and capacitance intuitively.
Starting with definitions,
$$C = \frac{Q}{V_{applied}}$$
$$C = \epsilon\frac{A}{l}$$
the second one is for plate capacitors.
To my knowledge, when an electric field applied, within a material there will be a charge separation.
Eventually, there will be two opposing forces on every charged particle: one from the applied field and the other is from the induced field due to the charge separation.
At equilibrium, those two forces are balanced and canceling each other thus a net force on a charge will be zero.
Mathematically, for the charge q
$$ qE_{applied} + qE_{induced} = 0$$
$$ |E_{applied}| = |E_{induced}|$$
By using the Coulomb force equation, I suppose that
$$ |E_{applied}| = |E_{induced}| = \frac{Q}{4\pi\epsilon{d^{2}}}$$
where $d$ is the distance between the charge pair.
Eventually, I can get
$$ 4\pi\epsilon{d} = \frac{Q}{E_{applied}d} = \frac{Q}{V_{applied}}$$
as you can see, this kind of relationship is similar to the definition of capacitance;
$$C = {\frac{Q}{V}} = \epsilon\frac{A}{l}$$
So my question is, first, do I correctly derive (qualitatively) the relation between the permittivity and the capacitance?
Second, the permittivity sounds like to me that it is a degree about how an electric field is permeable through the space (like within a material).
But I am not sure how this property is directly connected to the number of separated charges like in the definition of capacitance.
 A: I will answer the second ("intuitive") part of your question, for the case where the frequency of the voltage signal applied to the plates of the capacitor is low.
As you point out, for a capacitor in air, the capacitance C is equal to the permittivity of air times the plate area divided by the gap distance between the plates.
This means that if we make the gap smaller, the capacitance for a given area goes up. This lets us make capacitors with large C, without making them physically large, by using a small gap.  But the closer the plates get to one another, the more risk there is of having the air in the gap stop being an insulator- and having the capacitor discharge itself via a spark in the gap.
Now we note that if we fill the gap with a thin sheet of an insulating material which contains polar molecules that can be easily realigned with the application of an external electric field, the capacitance of the capacitor increases and its resistance to breakdown is improved, since the air is gone.
Intuitively, the presence of "dielectric" (polarizable and insulating) material in the gap has the effect of tricking the two plates into behaving as if they were closer together than they actually are.
