What is electric flux density?

Question

We say the electric flux is the number of field "lines", thus electric flux density is the number of field "lines" per a given area. However, let's say we had a point charge $Q$ centered at the origin and we were to enclose this charge with a surface of radius $R$. If we wanted to integrate the electric flux density over that surface, why do we get the total charge enclosed instead of the electric flux? I know this mathematically makes sense because we can write Gauss's law and multiply both sides by epsilon to yield this result, but I don't think this makes sense with what an electric flux density "is". Do I have a misunderstanding of what an electric flux density "is"? Please keep the math simple as I am only in first year university with limited knowledge of multivariate calculus.

\begin{align} \int \vec{E} \cdot ds &= Q / \varepsilon \rightarrow \varepsilon \int \vec E \cdot ds = Q_\text{enclosed} \\ \int \vec D \cdot ds &= Q_\text{enclosed} \qquad \vec D = \varepsilon \vec E \end{align}

Answer 1

It seems that you're likely to be using non-standard terminology.

If I understand your comments correctly, you seem to be associating the name "electric flux density" with the vector field $\vec D= \varepsilon \vec E$. If that is the case, then this is definitely non-standard terminology. The $\vec D$ field certainly does suffer from a lack of a universally-accepted name (though the problem is less bad than in magnetism), but a better name is the one used in Wikipedia, "electric displacement field". Using the name "electric flux density" for $\vec D$ is acceptable when writing in a generic context, but it is vital that it be clearly identified as such when it is introduced.

To be clear: the electric flux across a surface $S$ is defined as $$ \Phi(S) = \iint_S \vec E(\vec r) \cdot\mathrm d\vec S, $$ in terms of the electric field $\vec E$ itself. If the surface $S$ is closed, then the electric flux is still $$ \Phi(S) = \oint_S \vec E(\vec r) \cdot\mathrm d\vec S $$ and it just happens, because of Gauss's law, to coincide with $Q_S/\varepsilon_0$, i.e. the total charge enclosed by $S$ divided by the vacuum permittivity. Moreover, if you're dealing with the electric field $\vec E$ produced by a free-charge distribution that's embedded in a homogeneous, isotropic linear dielectric with permittivity $\varepsilon$, then $\Phi(S)$ can also be seen to equal $Q_{S,\mathrm{free}}/\varepsilon$, where $Q_{S,\mathrm{free}}$ is the free charge enclosed by $S$.

Because of this, the direct understanding of the term "electric field density" is to assign that to the vector field $\vec E$ itself, since it is the vector field whose surface integrals give the electric flux.

It is possible to start re-defining those terms so that you have a bit more operational room in how you distinguish between $\vec E$ and $\vec D$, though you run the risk of creating an extremely confusing situation. It looks to me that what's happened is that your lecturer has chosen a confusing choice of terminology and that's led you into some conceptual contradictions which are impossible to clear up within that framework. However, without seeing the notes in full as provided directly by your lecturer, it's impossible to tell that for sure.

Answer 2

It's just proportional to the electric field in order to remove the epsilon constant. There cannot be any other 'intuition' that doesn't also apply to electric field because the constant of proportionally is somewhat arbitrary.

I think there's an issue with your question:

Why do we get charge enclosed 'instead of' electric flux?

Well Gauss's law says that the two are equal for a closed surface (by a constant factor). The explanation of Gauss's law has probably been covered on this website