Why isn't the electrical field between two parallel conducting plates quadrupled? For the sake of clarity and brevity, here is the description from my textbook (Halliday & Resnick 9th):

This description states that the electric field is doubled. I am seriously confused. When the two plates are brought together, shouldn't the field quadruple? The surface density doubles on each inner surface, doubling their generated electric field. Additionally, these two fields should obey superposition rules of vectors, doubling the net-field, for a quadrupled total?
 A: I personally think the text is misleading. It's blindly applying Gauss' law while not considering its subtleties.
Here's a more cause-and-effect way to look at it. After this, we'll get to Gauss' law.
Let's take a look at the positively charged plate. Yes, the surface charge density on one side doubles. But the surface charge density on the other side goes to zero. A known result about infinite sheets of charge is that distance from the sheet doesn't affect the electric field. So, in effect, the region to the right of the positive plate is not affected by the redistribution of charge on this plate. The total surface charge density (if you were to flatten the plate and look at it as truly two dimensional) didn't change.
So, the positive plate's effect on the electric field isn't changed when the negatively charged plate is brought near. Instead, it's the presence of the second negatively charged plate that doubles the electric field in the region between the plates, not the doubling of the surface charge density on one of the plates. There is only one doubling going on.
In fact, if you were to somehow take two infinite sheets each with $\sigma$, and separated them by some small distance, the electric field outside of the region between the plates would be identical to a single infinite sheet with $2\sigma$. This is a crucial idea. I hope it's clear.
Why, then, is the text saying it has to do with the surface charge density?
They are applying Gauss' law to a Gaussian pillbox with the left face in the middle of the positive conductor and the right face in the vacuum between the plates. 

With this choice of Gaussian surface, there is only a flux through the right face. This is convenient for calculations. However, the electric field that gives rise to this flux is not due only to the enclosed charge, even though we mathematically calculate it like that. Rather, the electric field one should use in Gauss' law is the total electric field due to all charge distributions, including charges outside the Gaussian surface. This is one of the subtle but amazing facts of Gauss' law: To calculate the flux through a surface, you only need to mentally worry about the enclosed charge, but the result you get for the electric field (if you can indeed extract the field from the flux, usually in a highly symmetric geometry) is due to all of the charge, not just the enclosed charge.
So, for your particular problem from H&R, the flux through the right face of the Gaussian surface is caused by the superposed electric field from both plates. Neither of these fields alone changed, but their effects superpose, causing a doubling of the net electric field. But when applying Gauss' law for this problem, we usually don't worry about what actually causes this electric field. The doubling of the electric field is "accounted for" mathematically by the doubling of the surface charge density on the positive plate. However, viewing this doubling of the charge as the cause of the now-doubled electric field is not correct in my opinion.
A: Since the plate is a conductor the surface charge is spread across both sides of plate (see fig 23.16 a or b). Hence this gives that the electric field in between the two plates is actually 
$$E=\frac{\sigma}{2\epsilon_{0}}$$
This comes from Gauss' Law $\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enclosed}}{\epsilon_{0}}$
Now, the electric field for two capacitors of $\sigma$ surface charge has an electric field inside of 
$$E_{1}=\frac{\sigma}{2\epsilon_{0}}-\left(\frac{-\sigma}{2\epsilon_{0}}\right)=\frac{\sigma}{\epsilon_{0}}$$
So now if we double the surface charge we get
$$E_{2}=\frac{2\sigma}{2\epsilon_{0}}-\left(\frac{-2\sigma}{2\epsilon_{0}}\right)=\frac{2\sigma}{\epsilon_{0}}=2E_{1}$$
A: I have doubts about only one conducting plate. H&R say that the field outside is E=σ/2ε0, but if you add vectorially the two fields due to the two surfaces, then the field is E= σ/ε0. Inside the plate the vector addition is zero. 
When there is two plates the field inside the metal of the two plates, due to the four surfaces, is not zero and there must be a redistribution of charge. One can make a drawing of the vectors. Outside of the two plates the sum of the four fields is zero. To calculate the total field between the plates, I think we must take into account only one time the field E= σ/ε0 because the same line of field that leaves one positive charge in one inner surface, enters in other negative charge in the other inner surface. Then the field between the plates is E= σ/ε0
