In this example of Griffiths, we see a cylinder with the given volume charge density. It turns out that the electric field only goes outward in the direction perpendicular to the curved surface, so in the $\hat{s}$-direction / perpendicular to the axis of the cylinder.
However, why is there exactly no electric field in the direction going parallel to the axis of the cylinder / outward perpendicular to the dark gray flat disk on the left edge (and perpendicular to the flat disk on the right edge)?
Since the electric field of this cylinder is basically the flat dark gray disks superimposed onto each other. Then the field lines going outward perpendicular to the flat dark gray disk wouldn't be cancelled right?
Is this field just ignored for some reason since it is a long cylinder? How do you know that it will be negligible in that case?
Best Answer
You are assuming that the cylinder is long and that you are measuring the field "somewhere in the middle," where the fields either cancel out or are sufficiently far away that they're negligible.
As somebody who is currently going through Griffiths, I honestly wouldn't worry about it much. He presents more rigorous ways of calculating the electric field. You can use those and the appropriate maths (cylinder length approaches infinity, numerical techniques, etc.) to convince yourself that the field only depends on $\hat s$ (to a first degree approximation when the cylinder is finite).