Can you elaborate the concept of symmetry and how do you conclude that the field is $0$ along $x$-$z$ plane? This is a fairly popular problem.

An infinite slab, of thickness 2d, carries a volume charge density $\rho$.Find the electric field, as a function of $y$, where $y=0$ at the center. Plot $E$ versus $y$ calling $E$ positive when it points in the $+y$ direction and negative when it points in the -y direction.
  

Why is the field along the $x$-$z$ plane $0$? The solution says "by symmetry".
 A: If you take any point on the $XZ$ plane and calculate the electric field due to a point on the slab, there will always be precisely one point in the exact opposite direction, exactly the same distance away (as in taking the negative of the displacement vector) also on the slab. This other point will also produce an electric field at the point. As it is the same distance away it will be of the same magnitude as the first point and as it is in the opposite direction, the electric fields will cancel out.
This applies to every point - each one has an "opposite" on the other side of your point that will also exhibit the same electric field, but in the opposite direction. Therefore each point on the slab can be "paired up" and you will never get an electric field at any position on the $XZ$ plane.
Note that this argument only works because the slab is infinite and symmetrical in the $XZ$ plane; hence the phrase "by symmetry".
A: I thought it wasn't defined in $ y = 0 $, but here is my argument anyways. 
In general, bilateral symetry of a charge configuration in a plane $ x = k $, implies that the field has no $\hat{x}$ component on $ x = k $.
In the charged plane example, you have bilateral symetry for all $ z = k_1 $ and $ x = k_2 $ (that's why the field has only $\hat{y}$ component in all points), but also in $ y = 0 $. As we saw, that imples that, at $ y = 0 $, field has no $\hat{y} $ component either. Then, the field must be $\vec{0}$. 
