As shown in the figure, two large conducting plates are positioned...

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

The electric field produced by a large plane sheet of charge with surface charge density $\sigma$ is given by $E = \dfrac{\sigma}{2\varepsilon_0}$.

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

The electric field due to the negative plate is directed towards the right (towards the plate) with magnitude $E_- = \dfrac{\sigma}{2\varepsilon_0}$.

Since these two fields are in opposite directions, the net electric field is their vector sum:
$E_{net} = E_+ - E_- = \dfrac{\sigma}{2\varepsilon_0} - \dfrac{\sigma}{2\varepsilon_0} = 0$

Therefore, the electric field to the left of the plates is zero.