How is electric flux related to permittivity? How is Gauss' law related to permittivity?
I know that it equals $1/\epsilon_0$ times the magnitude of the charge enclosed. But, I'm unable to understand what it actually means.
Can someone intuitively explain it to me?
 A: Wikipedia says

In electromagnetism, absolute permittivity is the measure of the
resistance that is encountered when forming an electric field in a
medium. In other words, permittivity is a measure of how an electric
field affects, and is affected by, a dielectric medium. The
permittivity of a medium describes how much electric field (more
correctly, flux) is 'generated' per unit charge in that medium.

It is clear that more the permittivity of a matter, more resistance is offered to the electric field  resulting in decrease if the flux in Gauss law. In other words
Flux in inversely proportional to permittivity.
You can visualize it as, suppose 10 magnetic field line are produced due to the charge but due to the dielectric matter the number of field lines passing through the Gaussian surface decrease and the decrease is proportional it's permittivity.
You can visualize entire electrostatics on electric field lines, this will help you a lot to grab the concepts.
A: I'll consider only a vacuum (free space). I'm also not sure if this is the 'intuitive' answer you seek, but it's how I think of it.
One can view permittivity $\epsilon_o$ in Gauss' law
$$\Phi_E\equiv\int \vec{E}\cdot d\vec{A}=\frac{Q_\text{enclosed}}{\epsilon_o}\tag{Integral form of Gauss' law}$$
as playing the role of a constant of proportionality between the charge enclosed $Q_\text{enclosed}$ and electric flux $\Phi_E$. That is one could deduce that flux and the enclosed charge as being proportional to one another:
$$\Phi_E\propto Q_\text{enclosed}.$$
The above proportionality means that increasing the enclosed charge by a factor of, say $n$, would cause an increase in electric flux by the same factor. Now to make it a real equation, we just add a constant that makes the units and numerical values agree on each side of the equation:
$$\Phi_E=AQ_\text{enclosed}.$$
It happens to be more convenient to write the constant $A=1/\epsilon_o$.
So, in a vacuum, one can view permittivity as playing the role of a "conversion factor" between flux and enclosed charge.
