EM waves griffith

  • Imagine you are sipping tea or coffee while discussing various issues with a broad and diverse network of students, colleagues, and friends brought together by the common bond of physics, graduate school, and the physics GRE.

Post Reply
blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

EM waves griffith

Post by blackcat007 » Wed Jul 08, 2009 1:34 am

in Griffith in the topic of boundary conditions where one of the medium is a conductor, he has taken Kf ie (the free surface current) to be 0, the reason he states is that for ohmic conductors (J=sigmaE) and says that for surface current to exist there should be an infinite electric field..
i don't understand this point.

can anyone explicate this point?

User avatar
quizivex
Posts: 1031
Joined: Tue Jan 09, 2007 6:13 am

Re: EM waves griffith

Post by quizivex » Wed Jul 08, 2009 2:26 am

(edited)

In the Ohm's law expression J = sigma * E, J is a volume current density, which is a current per unit area perpendicular to the flow direction (amps / m^2)

A surface current density is amps per unit length perpendicular to the flow. This 'length' is just a 1-D line... it has no area. Thus, to have a finite current K along a 2-D surface, J would be infinite, and by Ohm's law E would also have to be infinite to produce it. Since that would be non-physical, he sets K=0 in the boundary conditions.

blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

Re: EM waves griffith

Post by blackcat007 » Wed Jul 08, 2009 2:58 am

quizivex wrote:In the Ohm's law expression J = sigma * E, J is a volume current density (amps / m^3)

A surface is just a 2-D plane, so it has zero volume. Thus, to have a finite current along a 2-D surface, J would be infinite, and by Ohm's law E would also have to be infinite to produce it. Since that would be non-physical, he sets K=0 in the boundary conditions.
but J is dI/da
and also if i have free charge on the surface from prior, ie say i put some extra charge in the conducting region, then they would flow to the outer surface, so in that case due to the oscillating E field, there can be surface current isn't it?

blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

Re: EM waves griffith

Post by blackcat007 » Wed Jul 08, 2009 3:15 am

blackcat007 wrote:
quizivex wrote:In the Ohm's law expression J = sigma * E, J is a volume current density (amps / m^3)

A surface is just a 2-D plane, so it has zero volume. Thus, to have a finite current along a 2-D surface, J would be infinite, and by Ohm's law E would also have to be infinite to produce it. Since that would be non-physical, he sets K=0 in the boundary conditions.
but J is dI/da
and also if i have free charge on the surface from prior, ie say i put some extra charge in the conducting region, then they would flow to the outer surface, so in that case due to the oscillating E field, there can be surface current isn't it?
also in that case the normal component of the electric field in the non conducting region won't be 0 since sigmaf is not 0 now.

User avatar
quizivex
Posts: 1031
Joined: Tue Jan 09, 2007 6:13 am

Re: EM waves griffith

Post by quizivex » Wed Jul 08, 2009 3:55 am

blackcat007 wrote:but J is dI/da
Oops, yea I messed up the units, but the conclusion is similar. J is current per unit area perpendicular to the flow direction. They just call it a volume current because it flows in space as opposed to a surface or a line.

So for a current on a 2-D surface, the flow is in some direction, but the region perpendicular to that direction, within the surface, is just a 1-D line, which has no area, so J is infinite in a surface current.
blackcat007 wrote:and also if i have free charge on the surface from prior, ie say i put some extra charge in the conducting region, then they would flow to the outer surface, so in that case due to the oscillating E field, there can be surface current isn't it?
The net "free charge" in a conductor goes to the surface, but that has nothing to do with the current flow, which can still pass through the interior as a volume current density.

If you're trying to consider what would happen if you had free surface charge on a conductor and applied light or an E-field to it, I'm just guessing here, but I think that "ruins" the original problem because the conductivity is an intrinsic property of the metal, depending on factors such as the (volume) density of loose electrons. Once you add more electrons externally, you're actually changing sigma, so it's a new problem. I don't know for sure, but if you add a surface charge density I think sigma itself might become infinite on the surface. Maybe it leads to a simple answer, or maybe all kinds of weird *** happens, lol, but I give up here.



Post Reply