Reflection at a conducting surface

[phrygian]

In Griffiths Electrodynamics (page 397) electromagnetic wave reflection at a conducting surface is introduced. He says "for ohmic conductors, Jfree = sigma * E, there can be no free surface current since this would require an infinite electric field at the boundary."

I don't understand why this is so, from Jfree = sigma * E it seems a finite electric field would induce a finite free current, can anyone help me make sense of this?

Thanks for the help.

 

[Bill_K]

Since J_free = 0, it means that E = 0. Since the total E must be zero, E_refl = - E_inc.

 

[Meir Achuz]

G is a bit confusing here. Since his E_T in Eq. (9.142) is not zero, there must be a non- vanishing j at the surface. It is K, an idealized surface current, that does not exist for a finite conductivity. For a perfect conductor, for which G gives E in Eq. (9.148), the boundary condition on B requires a surface current K. This surface current is equivalent to the integral of j from zero to infinity (in depth) for the finite conductivity case.

 

[Meir Achuz]

Since J_free = 0, it means that E = 0. Since the total E must be zero, E_refl = - E_inc.

This is for the perfect conductor case, not what G discusses in Eqs. (9.139)-(9.147).
For that case j does not equal zero at the surface.
G does not say j=0 at the surface but that the "surface current" is zero,, by which he means K for the finite conductivity case.

 

[Antiphon]

This can be confusing.

Its the surface version of resistors and inductors.

An incoming planewave striking a surface of finite conductivity will have two sources of physical current induced on it.

One component of current is due to a finite non-zero resistivity of the surface. It is ohms law, J = sigma * E as above. In this situation there is a non-zero tangential electric field, and the current J is in the same direction. There is ohmic heating of the surface etc.; everything you'd expect of a "resistor".

There is also another component of current from induction via the magnetic field (H in MKS units.) It's magnitude is twice the incident magnetic field. For a perfect conductor, this electric current flows even though there 0 tangential electric field. It is the equivalent of the current induced in a wire (shorted) when it links time-changing flux. This current does not represent ohmic heating and power loss. The boundary condition here is J = 2n X Hi where n is the surface normal and Hi is the incident magetic field. This current is also in the direction of the (incident) electric field. The total electric field for this case is zero as mentioned.

 

[Meir Achuz]

There is only one current in each case. For a 'perfect conductor', there is only a surface current K. For a conductor with a finite conductivity, there is only a volume current density j.

 

[Antiphon]

There is only one current in each case. For a 'perfect conductor', there is only a surface current K. For a conductor with a finite conductivity, there is only a volume current density j.

You know that can't be correct by smoothly taking the surface resitivity from one ohm/square to zero. The inductive term doesn't abruptly turn on on at 0.

The volume is steadily and smoothly decreasing as the penetration depth goes to zero together with the resistivity.

 

[Meir Achuz]

An effective surface current K is given by the integral from zero to infinity in depth of the volume density j. j falls off like exp[-x/\delta].
If delta is large, surface current has little meaning because j is too spread out.
As the conductivity gets larger, delta approaches zero, like you say.
Then only K has meaning.

 

[Antiphon]

Ok, I misunderstood you. We were in agreement.

Yes, there is only one current.

 

[tapsanit]

How's about the surface charge density. Does it follow discussed idea so far? I mean there is induced surface charge density if and only if there is current flowing on the surface. But we can not expect the surface charge under the ohmic conductor, finite conductivity?

 

[Meir Achuz]

There can be a surface charge density for any conductivity.

 

[tapsanit]

So, in the case of ohmic conductor in which the conduction current plays the important row than the surface current, then the surface current could be vanished, the surface charge density should also be vanished, because the changing of surface charge density defines the surface current due to the conservation of charge.

How possible it is to have some surface charge without surface current if you are in the situation saying that there is just one current in each case.

The reason why i am asking this because I am considering the boundary condition between dielectric and conductor, thanks for every reply above.

 

[Born2bwire]

Surface charge and surface current are two separate things. One rough way to look at it is that the tangential electric field's boundary conditions help define the surface current while the normal electric field's boundary conditions help define the surface charge.

The surface charge represents a net charge density while the surface current represents a net flow of charges. You can certainly have a net surface current without a net surface charge (just think of an equal density of electrons flowing in one direction and an equal amount of positive charges flowing in the opposite direction). The idea that the surface charge is more related to the normal electric field is that you can very roughly picture the normal electric fields drawing up the charges in the bulk. But at the boundary, the charges are restricted and so they collect at the surface until the boundary conditions for the normal electric field are satisfied. On the other hand, the tangential electric fields will move charges along a path that will not see any discontinuities and so there are no "boundaries" where the charges will bunch up and collect.

 

[tapsanit]

I do agree with you in the sense that the normal component of electric field might be considered to generate the surface charge. Actually, I think we can obtain the solution of the normal component by dealing with just the boundary condition in tangential component without the surface current.

Unfortunately, I strongly do not agree with you in the concept of having surface current without the surface charge density as you have replied by considering the flowing of free electrons relative to the equally positive charge. If you think about the conductor with finite size, e.g., small finite slab or finite cylinder, once light is applied to the conductor, there is the polarization of cloud of electrons and positive core, this mean that there are three regions; neutral region (no net charge), negative charge region (net negative charge of electrons) and positive charge region(net positive charge of positive core). This leads to the surface charge density depending on the position on the surface of the conductor.
Of cause, if you sum up two regions of equally opposite charge, you get zero net surface charge just like in the case of electrostatic.

 

[tapsanit]

So, what I think is the surface charge density could be generated by the polarization of cloud electrons relative to the positive core in conductor. This indicates that even no having the surface current, we can have the surface charge density according to "clem".

 

[Meir Achuz]

Tapsanit:

Born2bwire gave a complete answer in post #13. Try to understand it.
Most of your post #14 is wrong.

 

[tapsanit]

clem :

Could you describe why it is wrong, or it is just wrong because it dose not satisfy Born2bwire.

 

[tapsanit]

I asked the first question that if there was no surface current, there would not have the surrface charge density? The answer were,there was the surface charge density, and you can calculate it from the boundary condition of normal conponent, so perhap the origin of the surface charge density come from the normal component of electric field? Another answer, there can be no net surface charge density while the current is flowing on the surface showing independent of surface current and curface charge density

What I replied what the reson how the surface charge is generated.
Could you explain more why my opinion is not correct.

 

[tapsanit]

Lets me show the example, the metallic sphere is embeded within the vacuum, it is applied by the static electric field amplitude E0 polarized in z direction . Assume the perfect conductor, there is just the induce the surface charge generating the internal electric field to cancel the external field. This induced surface charge density is calculated from the normal component of electric field, and it is simply 3*eps0*E0*cos(theta). This amount of surface charge can be considered from the polarization of two spheres of different charge.

 

[Born2bwire]

Again, surface current and surface charge are two different things. This has to be true in some respects when we consider the steady-state situation due to the finite source of charges. If we take some generic volume element inside some larger bulk material, we can define the charge of the cube to be the sum of the positive and negative charges while the current is the charge flux through any of the surfaces. If we were to consider charge and current to be irrevocably intertwined, then a current running through the cube would have to relate to the charge. But what happens if the current entering and leaving the cube is the same (no net flux)? Then the amount of charge inside the cube does not change and so the current has no effect on the amount of charge inside (or on) the cube. Only when there exists a flux imbalance (represented as a net flux over all surfaces) does the current relate to the charge by virtue of the fact that now the current is either sinking or sourcing charges from the cube. But in steady-state this is untenable by the fact that it would require the ability to draw an infinite amount of charges into the volume or be able to contain an infinite amount of charges inside the volume as time progresses.

What happens at a surface is that if we choose a similar volume element that shares part of the boundary then we know that the surface along the boundary cannot have any charge flux (lest the charges be flying off the object). So if we induce an electric field normal to the surface then there is a first order force acting on charges to bring them into the volume element in the form of a current. But because they cannot flow out of the volume element, they collect along the surface until the second order electric field produced by the net charge density cancels out the applied normal electric field.

However, a tangentially applied electric field exerts a first order force on the charges to flow through the volume element. But since they are not impeded by a surface they can pass in and out of the volume element. Thus, no steady-state collection of charge can occur when we consider an infinitesimal volume element in this manner.

 

[tapsanit]

Thank you for your replying, I think this is very good answer applied to the case of conductor.

 

[aperception]

Sorry to revive this thread again, but I'm puzzling over it myself, and I'm still not convinced by any of the answers to the OP question - what exactly does Griffiths mean by "this would require an infinite electric field at the boundary"? What are the relevant equations to see that this is true?

I understand that he is talking about K rather than J, and the argument given by Clem that K is meaningless for finite conductivity because the skin-depth is non-zero is plausible, though even there I don't see why there can't be both a K and a J. Is it really true that K must be the integral of J over depth? I don't remember seeing this anywhere in Griffiths.

Any insight would be appreciated.

 

[Meir Achuz]

Griffiths tries to simplify, but this is not that simple. You might try a graduate treatment like Sec.11.2.3 of Franklin, 'Classical Electromagnetsim', but it is a bit complicated.
K would exist for a perfect conductor, but is an abstraction for a real conductor, which only has j. No conductor has both j and K.

 

[aperception]

OK thanks for your reply, I guess I will take your word for it (that there can't be both J and K in a conductor) until I get around to studying E&M at graduate level.