Electric field due to a finite and infinite sheet of charge.

[Badger01]

Can some one explain why the value for the electric field strength is $E = 2\sigma/\epsilon_0$ for a charged sheet but $E = \sigma/\epsilon_0$ if the sheet is infinitely large?

I understand the procedure of using Gauss' law and creating a Gaussian surface of constant E; however, I don't understand why the area of the Gaussian surface =2A in one case and =A in the other.

 

[Lavabug]

I think your 2 belongs in the denominator, that expression corresponds to the electric field of a single charged plate (provided it is infinitely big, or you are far away enough from the boundaries that the expression is accurate).

The electric field between two charged plates of equal dimensions with the same charge density would be the sum of the contributions of both plates, so you'd get twice that amount (canceling your 2 in the denominator) giving you sigma/epsion.

 

[Badger01]

Ahh ok, thanks. I can at least see how you derive the first from the second now, thanks. But I'm still not sure why you must take into account both sides of the plate for and infinite plate.
[and yes you're right the 2 should be on the bottom in the first expression]

 

[Lavabug]

Here's your answer: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

 

[phyphenomenon]

I also have the same question. In the finite sheet, we consider a Gaussian surface on only one side of the sheet. In the case of an infinite sheet, we consider two sides for Gaussian surface. Why? I couldn't catch this idea.

 

[Doc Al]

I also have the same question. In the finite sheet, we consider a Gaussian surface on only one side of the sheet. In the case of an infinite sheet, we consider two sides for Gaussian surface. Why? I couldn't catch this idea.

It has nothing to do with finite versus infinite sheets. But it does have to do with whether you're talking about the surface charge on a conductor or just a sheet of charge. Read the link that Lavabug posted almost two years ago.

 

[phyphenomenon]

Yes. Thank you. I have another question. Is Gauss law applicable to only uniform charge distribution? On formulating Gauss law, textbooks consider a closed surface S in a non uniform field.

 

[Doc Al]

Is Gauss law applicable to only uniform charge distribution?

No. It applies to any charge distribution.

 

[Deepak verma]

you have to draw a function w.r.t to the distance from the center , integrate it up to the distance till the charge distribution is mentioned along its area or volume whatever charge density is given , result would be the total charge contained by that object and you can use gauss theorem after that.