Answer
$E=0$
Work Step by Step
According to Equation (23-13), the magnitude of the electric field due to the plate of positive charge is $E = \frac{\sigma}{2~\epsilon_0}$ and this electric field is directed away from the positively charged plate.
According to Equation (23-13), the magnitude of the electric field due to the plate of negative charge is $E = \frac{\vert \sigma \vert}{2~\epsilon_0}$ and this electric field is directed toward the negatively charged plate.
To the right of the plates, the electric field due to the negatively charged plate is directed to the left while the electric field due to the positively charged plate is directed to the right. Since the magnitude of the surface charge density is equal for both plates, the magnitude of each electric field due to each plate is equal.
Therefore, by superposition, $E = 0$
