How is dielectric constant both $E_{net}/E_o$ and $ε/ε_o$? The questions asks that, given the (surface) charge density (σ) on the plates of a parallel plate capacitor and dielectric constant (k), what would be the magnitude of induced charge density on the ends of the dielectric slab inserted in between?
If I use both facts that k = Enet/Eo and k = ε/εo in the same question, I get the wrong answer - the correct answer is σ_ind = σ(k-1)/k.
Here's my attempt at the problem posted in handwritten form, hopefully it's clear enough:

However, I'm aware that the actual formula for this has a 'k' factor in the denominator on the RHS. Could somebody point out where am I going wrong in this particular derivation?
The point of me asking this question is to indirectly address the fact that I'm not comfortable that k is both Enet/E0 and ε/ε0. I can't derive one from the other. That's why I used both to prove that I'm arriving at a false result.
 A: You are wrong in writing $E_i = \frac{\sigma_i}{\epsilon}$
The correct way would be $E_i = \frac{\sigma_i}{\epsilon_0}$.
You see, the reason we want to consider the dielectric constant is that we don't want to work with the induced charges. Using the dielectric constant, we can just see the free charge (the on present on the plates here), and find out what will be the net field inside the dielectric. Why that happens is because the induced charges produce an opposing field. But to find the field due to the induced charges you don't need to consider that it is a dielectric, because then you'd be overcompensating for that.
Edit after comment:

I didn't understand your point properly, but shouldn't it matter what material is b/w the induced charges? Let's ignore the fact that maybe we're overcompensating by introducing the same fact twice in the derivation (although I don't quite understand what you meant by that). If I were to derive the electric field formula Ei=σi/ϵ using Gauss' Law from scratch, I would have to use ϵ instead of ϵo and it would carry forward to this statement. Is that wrong to say so?

By Gauss' law you will get $E = \frac{q_{total}}{\epsilon_0}$, and not $\epsilon$. The formula I think you are referring to is the version of Gauss' law when we have dielectrics, which states that $E = \frac{q_{free}}{\epsilon}$, where $q_{free}$ is the free charge (the one we place on the plates here, and not the induced charge). You cannot use this directly to calculate the field of the induced charge, because this formula was derived keeping in mind how much extra induced charge will induced.
Also when you are trying to find the electric field due to the induced charge it won't matter what dielectric is present inside (except in determining the final electric field which will be $\frac{E_0}{k}$). The usual Gauss' law and Coulomb's law can be used without any kind of correction. When we do make corrections to these laws in the presence of a dielectric it is because we want the electric field by just knowing the free charge and we don't want to consider the induced charge in our calculation. This is also what I meant by overcompensating. When you divide by $epsilon$, you are automatically considering the extra field that will be produced by the charge induced by to the electric field we are applying. But here it is the induced charge whose field you are calculating. This induced charge's field won't induce any other charge.
If this doesn't clear your doubt, please make another comment.
Edit 2:

I guess then my follow-up question will be then, that what is the significance of $\epsilon_0$ which we use in Coloumb's Law and Gauss' Law. Does it have a special meaning or are we just supposed to treat it like a constant? My previous notion was based on the observation that we use ϵ instead of ϵ0 in Coloumb's Law, leading me to believe that we tweak this constant depending on the material inside. But if I'm understanding you correctly, that is only the case because we just use the final electric field in the dielectric. The individual fields because of the charges, and because of the material would still be the same, i.e use $\epsilon_0$. It is only to directly calculate the final field, do we use $\epsilon$ or $k\epsilon_0$.

Yes you understood me correctly. As for your question, I don't know what you want to say by "special meaning", but yes, $\epsilon_0$ is a fundamental constant that is fixed no matter what. Like you said, it is only when we want to find the electric field inside the dielectric without worrying about the induced charges, do we use $\epsilon$ (also, even this can only be done when we have what is called a linear dielectric, which we usually do but not always). Using $\epsilon$ is a shortcut. If you use the usual Coulomb's law, using $\epsilon_0$, but considering all the induced charges as well, you will get the correct the answers.
