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In this answer by David Z, we can read,

When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes into account all the charge on both these surfaces, so it is the sum of the two charge densities.

$$\sigma = \frac{Q}{A} = \sigma_\text{inside} + \sigma_\text{outside}$$

With this definition, the equation we get from Gauss's law is

$$E_\text{inside} + E_\text{outside} = \frac{\sigma}{\epsilon_0}$$

where "inside" and "outside" designate the regions on opposite sides of the plate. For an isolated plate, $E_\text{inside} = E_\text{outside}$ and thus the electric field is everywhere $\frac{\sigma}{2\epsilon_0}$.

Now, if another, oppositely charge plate is brought nearby to form a parallel plate capacitor, the electric field in the outside region (A in the images below) will fall to essentially zero, and that means

$$E_\text{inside} = \frac{\sigma}{\epsilon_0}$$

There are two ways to explain this:

  • The simple explanation is that in the outside region, the electric fields from the two plates cancel out. This explanation, which is often presented in introductory textbooks, assumes that the internal structure of the plates can be ignored (i.e. infinitely thin plates) and exploits the principle of superposition.

    electric fields in superposition

  • The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own.

    electric field from one plate to the other

Either way, it's not true that $\lim_{d\to 0} E = \frac{2\sigma}{\epsilon_0}$.

I made the part I couldn't see the reason for in bold.

Let's call the charge density before bringing the other plate $\sigma$. The new charge density after bringing the other plate is $\sigma'=2\sigma$. Why can't we superimpose the two fields in the second scenario?

If we have a positive charge near a negative one, here too the negative charge acts as a "terminator", yet the total field is the sum of the fields produced by the two charges, not just one. enter image description here

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    $\begingroup$ IMHO, the quoted explanation is unnecessarily confusing and actually incorrect. I think your intuition in this case is a better guide. $\endgroup$ – S. McGrew Mar 4 at 17:20
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I agree with your assessment. The mistake David makes is assuming that since all of the charge moves to one side of the plate that the field on one side of the plate doubles and the field on the other side goes to $0$ due to that plate. Even when the charges all move to one side, the field on each side of the plate is still $\sigma/2\epsilon_0$. Then you would still add the two fields together.

The reason the offset of the charges doesn't change anything is because we have assumed infinite sheets of charge so that the field does not depend on the distance from them. Therefore, suggesting that the horizontal translation of charges produces different fields that do not superimpose appears to be incorrect. David's explanation is correct if you have already done the superposition (then you do have $0$ field outside and $\sigma/\epsilon_0$ inside with field lines starting on positive charges and ending on negative charges).

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