### Solution : Crossing a charged layer - theorem of gauss

1. Field .
Being a point M belonging to the Ox axis. The planes passing through this axis are planes of symmetry of the charge distribution, the direction of the electrostatic field is contained in all these plans: the direction of the field is carried by the axis Ox.
On the other hand, the distribution is invariant for all translation parallel to the plane yOz.
Therefore :
The yOz plan is also a plane of symmetry of the charge distribution. Then Ex (- x) = - Ex (x).
To determine the electric field, we apply the Gauss theorem considering as Gaussian surface  a cylinder of height 2x, with axis Ox and O its center .

2. Speed
A charge q of the same charge of the layer is initially blocked by the layer to the point of abscissa
x = 0. To cross this layer it is sufficient that the charge q reaches the plane yOz since beyond this level, it is pushed to the right.
Applying the theorem of kinetic energy to the particle between the abscissa point x = - a and x = 0.
We get :