Robert Koslover
Antennas, Waveguides, Electromagnetics
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Posted:
1 decade ago
2010年9月16日 GMT+8 上午4:37

Short answer: Well, for a *conducting* surface in empty space, the surface charge density can be expressed as the norm of the electric field divided by epsilon_0. I.e., surface charge density [coulombs/m^2] = normE_rfw/epsilon0_rfw. This follows simply from Gauss' law of electrostatics (which is valid in the RF case as well). (Remember: "Del dot E = rho / epsilon_not" ) (Just consider a thin pillbox around the surface and the integral form of Gauss' law.)

More complicated example: For an interface between two dielectrics, you''ll probably want to begin with D instead of E, and derive an expression that is similar, but based on the more general expression, "Del dot D = rho free" and also recognize the existence of electric displacement fields on *both* sides of the interface. For example, suppose your interface is the x-y plane, so that the normal points along the z axis. Then the free surface charge density on the interface should be given by Dz_rfw (on the right side) - Dz_rfw (on the left side). Note the discontinuity of the field, which I suppose may complicate this. Regardless, I suppose you could find the surface charge at a dielectric interface by interrogating your computed D-field results near the boundary on both sides, in post processing inspection. And then, of course, there is the "bound" charge density to consider also in the case of dielectrics. You might want to re-check the exact EM definitions, when working with dielectrics. Good luck.

Short answer: Well, for a *conducting* surface in empty space, the surface charge density can be expressed as the norm of the electric field divided by epsilon_0. I.e., surface charge density [coulombs/m^2] = normE_rfw/epsilon0_rfw. This follows simply from Gauss' law of electrostatics (which is valid in the RF case as well). (Remember: "Del dot E = rho / epsilon_not" ) (Just consider a thin pillbox around the surface and the integral form of Gauss' law.)
More complicated example: For an interface between two dielectrics, you''ll probably want to begin with D instead of E, and derive an expression that is similar, but based on the more general expression, "Del dot D = rho free" and also recognize the existence of electric displacement fields on *both* sides of the interface. For example, suppose your interface is the x-y plane, so that the normal points along the z axis. Then the free surface charge density on the interface should be given by Dz_rfw (on the right side) - Dz_rfw (on the left side). Note the discontinuity of the field, which I suppose may complicate this. Regardless, I suppose you could find the surface charge at a dielectric interface by interrogating your computed D-field results near the boundary on both sides, in post processing inspection. And then, of course, there is the "bound" charge density to consider also in the case of dielectrics. You might want to re-check the exact EM definitions, when working with dielectrics. Good luck.

Robert Koslover
Antennas, Waveguides, Electromagnetics
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Posted:
1 decade ago
2010年9月16日 GMT+8 上午4:40

Correction for the conducting surface case: Surface charge density = epsilon0_rfw*normE_rfw.

Sorry about that.

Correction for the conducting surface case: Surface charge density = epsilon0_rfw*normE_rfw.
Sorry about that.

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Posted:
1 decade ago
2010年9月17日 GMT+8 下午1:23

but "Surface charge density = epsilon0_rfw*normE_rfw" ,the normE_rfw means a scalar ,ie, normE_rfw is the magnitude of the electric field not a vector.

if the normE_rfw is normal to any surface ,then it could be right.

but "Surface charge density = epsilon0_rfw*normE_rfw" ,the normE_rfw means a scalar ,ie, normE_rfw is the magnitude of the electric field not a vector.
if the normE_rfw is normal to any surface ,then it could be right.

Robert Koslover
Antennas, Waveguides, Electromagnetics
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Posted:
1 decade ago
2010年9月18日 GMT+8 上午8:23

Well, I'm not 100% sure what you mean here, but you sound confused about "norm" vs "normal." You are correct that normE_rfw is a scalar. It is the magnitude (or "norm") of E. Now, for a perfectly conducting surface, E will also be perpendicular to that surface. And any vector perpendicular to a surface is said to be "normal" to that surface. So the vector E will be normal to the conducting surface.

And finally, in the case of a dielectric junction, although E (and likewise D) may not be normal to the surface, the normal (aka, perpendicular to the surface component) of E (or D) is still what you would want to use when extracting surface charge densities.

Does that help?

Well, I'm not 100% sure what you mean here, but you sound confused about "norm" vs "normal." You are correct that normE_rfw is a scalar. It is the magnitude (or "norm") of E. Now, for a perfectly conducting surface, E will also be perpendicular to that surface. And any vector perpendicular to a surface is said to be "normal" to that surface. So the vector E will be normal to the conducting surface.
And finally, in the case of a dielectric junction, although E (and likewise D) may not be normal to the surface, the normal (aka, perpendicular to the surface component) of E (or D) is still what you would want to use when extracting surface charge densities.
Does that help?