The answer by @BrianMoths is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation.
Start with your charge distribution and a "guess" for the direction of the electric field.
As you can see, I made the guess have a component upward. We'll see shortly why this leads to a contradiction.
Now do a "symmetry operation," which is a fancy phrase for "do something that leaves something else unchanged. In this case, I'm going to reflect everything about a horizontal line. I mean everything.
The "top" of the sheet became the "bottom." This is just arbitrary labeling so you can tell I flipped the charge distribution. The electric field is flipped too. (Imagine looking at everything in a mirror, and you'll realize why things are flipped the way they are.)
Hopefully, everything is okay so far. But now compare the original situation with the new inverted one.
You have exactly the same charge distribution. You can't tell that I flipped it, except for my arbitrary labeling. But if you have the same charge distribution, you ought to also have the same electric field. As you can see, this is not the case, which means I made a mistake somewhere.
The only direction for the electric field that does not lead to this contradiction is perpendicular to the sheet of charge.
for an infinite sheet of constant charge, the text says that the electric field is constant on any one side of the sheet. But that seems intuitively wrong to me, since I would think the field should be stronger the closer a point is to the sheet.
There's a geometric scaling argument at hand, and you probably need to appreciate Gauss' Law to get a real sense of it. It follows the same thinking that the $1/r^2$ law does.
In this argument for $1/r^2$ field strength from a point particle, it is seen that the solid angle that "A" occupies decreases at larger radii. If you consider a charged ball, then think about squishing it from a 3D object into a 2D piece of paper. That represents the area-based charge density.
An unsaid assumption for this line of thinking is that the field strength is determined by:
- The solid angle occupied by charged material
times
- The 2D charge density presented by that material
So extend this to an infinite sheet. No matter how close or far away it is, if you're looking parallel to the sheet, the sheet occupies exactly half of your field of vision. Furthermore, the surface charge density and angles of the sheet are also irrelevant of your normal distance.
Illustratively, you could apply this via the image above. Just remove the "A", and consider that this is an infinite sheet. As you expand the distance, you expand the area you sweep, but the angle times the charge density is invariant. Thus, the field is constant with distance.
Best Answer
Gauss' law relates the total electric flux leaving a closed surface to the charge enclosed in the surface. As you say, the electric field due to an infinite sheet of charge must, by symmetry, be perpendicular to the sheet (if it had a component parallel to the sheet, how could it prefer any direction over any other?). That means that for a cylindrical Gauss volume intersecting the sheet and with axis perpendicular to the sheet, the flux is entirely through the ends, with none through the sides. The enclosed charge is the same regardless of the length of the cylinder, so the flux through the ends must be also. Therefore the electric field is the same at any distance from the sheet.
This argument fails for a finite sheet, because the electric field is in general not perpendicular to the sheet, and so the total flux through the side of the cylinder is not zero.
To apply Gauss' law to an infinite line charge, use a cylinder whose axis coincides with the line. Then, by symmetry, there is no flux through the ends of the cylinder, and the flux per unit area through the tubular part is uniform. As you increase the radius of the cylinder, the enclosed charge is unchanged, so the total flux is unchanged. But the area of the tubular surface is proportional to the radius, so the flux PER UNIT AREA (i.e. the electric field) must be inversely proportional to the radius.