Let $$\sigma$$ be the uniform surface charge density of two infinite thin plane sheets as shown in the figure. Then the electric fields in three different regions, $$E_I$$, $$E_{II}$$ and $$E_{III}$$ are:

Let the unit vector $$\hat{n}$$ point from left to right (the positive direction). Both sheets have a positive surface charge density $$+\sigma$$.

For Region I (left of both sheets):

The electric field from the first sheet points to the left: $$\vec{E}_1 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second sheet points to the left: $$\vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_I = \vec{E}_1 + \vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n} - \frac{\sigma}{2\epsilon_0}\hat{n} = -\frac{\sigma}{\epsilon_0}\hat{n}$$

For Region II (between the sheets):

The electric field from the first (left) sheet points to the right: $$\vec{E}_1 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second (right) sheet points to the left: $$\vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_{II} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n} - \frac{\sigma}{2\epsilon_0}\hat{n} = 0$$

For Region III (right of both sheets):

The electric field from the first sheet points to the right: $$\vec{E}_1 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second sheet points to the right: $$\vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_{III} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n} + \frac{\sigma}{2\epsilon_0}\hat{n} = \frac{\sigma}{\epsilon_0}\hat{n}$$