Two large conducting plates are positioned parallel to each other...

Let the surface charge density of the positive plate be $+\sigma$ and the negative plate be $-\sigma$.

The electric field produced by a single large charged plate with surface charge density $\sigma$ at any point outside it is given by $E = \dfrac{\sigma}{2\varepsilon_0}$.

For a point to the right of both plates:
The electric field due to the positively charged plate (left plate) is directed away from it, i.e., towards the right, with magnitude $E_+ = \dfrac{\sigma}{2\varepsilon_0}$.

The electric field due to the negatively charged plate (right plate) is directed towards it, i.e., towards the left, with magnitude $E_- = \dfrac{\sigma}{2\varepsilon_0}$.

Since these two fields are equal in magnitude and opposite in direction, the net electric field to the right of the plates is:
$E_{net} = E_+ - E_- = \dfrac{\sigma}{2\varepsilon_0} - \dfrac{\sigma}{2\varepsilon_0} = 0$