Two large conducting plates are placed parallel to each other and...

Let the surface charge density on the inner surfaces of the positive and negative plates be $+\sigma$ and $-\sigma$, respectively.

The electric field at a point between the plates due to the positively charged plate is directed away from it and has a magnitude:
$E_+ = \dfrac{\sigma}{2\varepsilon_0}$

The electric field at the same point due to the negatively charged plate is directed towards it (which is the same direction as $E_+$) and has a magnitude:
$E_- = \dfrac{\sigma}{2\varepsilon_0}$

Since both fields point in the same direction (from the positive plate to the negative plate), the net electric field between the plates is the sum of their magnitudes:
$E_{net} = E_+ + E_- = \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0}$

Alternatively, using Gauss's law with a cylindrical Gaussian surface having one flat face inside the conducting plate (where $E=0$) and the other face in the region between the plates:
$\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{enclosed}}{\varepsilon_0}$
$E \cdot A = \dfrac{\sigma A}{\varepsilon_0}$
$E = \dfrac{\sigma}{\varepsilon_0}$