# Electromagnetic fields and waves 2

Discussion in 'Homework Help' started by mentaaal, Nov 2, 2009.

1. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Hey again guys (That means you steveb!)

Ok i been reviewing my fields and waves notes and have come across something i am not quite sure of.
It was proved that the electric field intensity E at a point p (in space above the plane) due to a uniform charge density ρs on an infinite plane is equal to ρs/2ε where ρs = surface charge density and ε is equal to the permittivity of the medium that p is in.

In this derivation there is no criterion required that states that the plane is a conductor or not as this is not required.

Reading over the boundary conditions it states that the normal component of the electric flux density vector Dn is equal to ρs when the Flux is coming out of a conductor. This is due to the fact that the tangential components of the electric field vector at the interface between two different media are equal. In a conductor, there is no tangential component so this is indirectly why D out of the conductor is normal.

So if Dn is = ρs then En is = En/ε
When this was stated there was no criteria given as to the dimensions of the conducting surface.

So looking back at the first example mentioned, if we assume that the infinite plane is a conductor, then is D = ρs/2 or ρs?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
Cool question. That one makes you think doesn't it. You've invented your own little paradox.

OK, the resolution of this paradox comes from understanding exactly what surface charge is. It is considered to be an infinitesimal thickness of charge that exists with a certain charge per unit area. In reality, the charges will exist over a few atom thickness, but for all practical purposes this is close enough for the macroscopic version of field equations that we normally use.

So, consider symmetry in this problem, and note that the question implies an infinitely thin plane. Whatever happens on one side of the plane, must also happen on the other side of the plane, but as a mirror image. This means that the space on one side of the plane sees Dn everywhere, and the space on the other side of the plane sees -Dn everywhere. Now apply the boundary condition and you get Dn+Dn=ps, which leads to Dn=ps/2

Your second case of the field above a conductor has the implication that the conductor is not infinitely thick, but substantially thick. In this case, the free charges in the conductor will move around and distribute themselves in whatever arrangement that allows the boundary conditions to be correct and results in zero electric field inside. Now, the only way for this to happen is in get a surface charge density that equals the Dn outside the conductor. Hence Dn=ps.

If you take the case of a conducting plate that is thin, but not infinitely so, then you get ps on one surface and ps on the other surface. The volume in between the two surfaces will not have a charge distribution. Now if you want to approximate this thin conductor as being infinitely thin, then you have to add both charge densities from the two surfaces, and you get 2*ps as the effective charge density of the infinitely thin conductor. ... hmmm ... Paradox solved!

3. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Thank you Jedi Steve! The force runs strongly in you!

P.S. come and lecture us deagnabbit!

Last edited: Nov 2, 2009