Electric Field Produced by an Infinite Sheet.

smithnh

I was browsing the internet looking for alternate proofs to why the electric field produced by an infinite sheet was equal to the charge density divided by twice the space permitivity constant. However, in my search I came across a proof that confused me at this link, http://www.physlink.com/Education/AskExperts/ae544.cfm

It found the electric field produced to be eqaul to the charge density divided by only the space permitivity constant. My question is, what happened to the factor of 2. Are they only considering one face of the sheet but even then should that not decrease it by another factor of 2. What is going on here?

ZapperZ

This is a common confusion if you do not look carefully at the nature of the charge distribution. On one hand, you have an infinite SHEET of charge, i.e. you have the charge occupying only a sheet, for example, the x-y plane. Here, the E-field is being shared in both spaces z>0 and z<0. So if you do Gauss's Law on here, the resulting E-field has a factor of 1/2.

On the other hand, if you have a conductor occupying the space at z<0, while the surface at z=0 (still the x-y plane) still has the charge, then a similar Gauss's Law application will get you E-field that is twice as large. All the E-field is in the z>0 space.

So just because you have an infinite plane of charge doesn't mean you can ignore the rest of the geometry of the problem.

Zz.

Doc Al

I didn't look at the proof in detail, but if they are talking about an ordinary sheet of charge and they left out that factor of 2, then they made an error somewhere.

(As Zapper points out, there's a common source of confusion when deriving the field from a charged conducting sheet compared to just a sheet of charge. But in that link I don't see them talking about conductors.)

Doc Al

They messed up the integration:

[tex]\int \frac{a}{(a^2 + x^2)^{3/2}} \;da= \frac{-1}{\sqrt{a^2 + x^2}} \neq \frac{-2}{\sqrt{a^2 + x^2}}[/tex]

smithnh

Thanks for the help.