Does Distance Affect the Electric Field Strength Above a Charged Plate?

AI Thread Summary
The discussion centers on the electric field strength above a charged plate, particularly questioning the relevance of distance in calculations. It is clarified that for an infinite sheet of charge, the electric field remains constant regardless of distance, meaning the field strength is the same at 0.1 meters and 1 meter above the plate. This contrasts with the electric field from a point charge, where distance is a critical factor, as described by the equation E = kq/r. The conversation highlights that while the electric field from a finite sheet may vary with distance, the behavior of an infinite sheet simplifies the situation. Overall, understanding the configuration of the charge is essential for accurately determining electric field strength.
oh.rry21
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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? :(
 
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oh.rry21 said:
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"
 
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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)
 
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