Gauss's Law and E fields

[oh.rry21]

So I'm trying to understand a couple things regarding this.

lets say we have a plate that's charged to Q. and we want to find the E field of a point .1 meters above the plate. when i looked at the solution, it never took the .1 meters into account when calculating the strength of the E field.

my roommate says that E field should be the same and thus the distance was irrelevant. so whether it was .1 meters or 1 meter above the plate, the E field would be the same.

but that doesn't make sense to me when you think about the equation for an electric field E= kq/r.

can anyone resolve that contradiction for me? :(

 

[Doc Al]

my roommate says that E field should be the same and thus the distance was irrelevant. so whether it was .1 meters or 1 meter above the plate, the E field would be the same.

He's right, at least if you're talking about the field from an infinite sheet of charge. Or a finite sheet of charge at distances much smaller than the dimensions of the sheet. Of course, if the sheet of charge is only 10 cm square, the field 10 m away certainly will depend on distance.

but that doesn't make sense to me when you think about the equation for an electric field E= kq/r.

That's the field from a point charge--a very different configuration from an infinite sheet of charge. (Of course, as you get far enough away from a finite sheet of charge the field begins to look more and more like the field from a point charge.)

You might want to browse through this site: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gaulaw.html#c4"

 

[buddingscientist]

Yep, because the e-field lines will be 100% perpendicular to the charged (supposedly infinitely wide) plate.
i.e. going straight up or straight down, so if you bound a gaussian cylinder above and through the plate, the field will go through only the circular discs at the end of the cylinder (and not the 'side walls' of the cylinder) - so it doesn't matter how high the cylinder is (drawing a picture would help heaps)