Assume a infinite depth good conductor block with width in y and length in x direction. Boundary is at z=0 and air is at z=-ve and conductor at z=+ve. Let a voltage apply across the length of the block in x direction so a current density established in +ve x direction. We want to look at the EM at the boundary just above and just below the surface.(adsbygoogle = window.adsbygoogle || []).push({});

1) We know at [itex] z=0^-[/itex](in air):

[tex]\tilde{E_-}(z) = \hat x E_0 e^{-\alpha z} e^{-j\beta z} \;\hbox { and }\; \tilde B = \frac 1 {\eta_0} \hat z \times \vec E_-(z)\;=\; \hat y \frac {\vec E_-(z)}{\eta_0}[/tex]

In air,His perpendicular toEonly because [itex]\eta_0[/itex] is real and don't have a phase shift.

2) We know tangential E continuous cross boundary therefore at [itex] z=0^+[/itex](in conductor):

[tex]\tilde{E_+}(z) = \hat x E_0 e^{-\alpha z} e^{-j\beta z} \;\hbox { and }\; \tilde B_+ = \frac 1 {\eta_c} \hat z \times \vec E_-(z) [/tex]

Here [itex]\eta_c[/itex] is complex and therefore has a phase shift, actually the phase angle is [itex]\theta=45^0[/itex]. This meanBis no longer perpendicular toEeven thoughBalso propagate in z direction.

The problem is the book still give:

[tex] \tilde B_+(z)=\hat y \frac {\tilde E_+(z)}{\eta_c}[/tex]

This meanBis still perpendicular toEinside the good conductor. I don't understand this. Please help me.

Thanks

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# EM wave right above and right below the conducting surface.

**Physics Forums | Science Articles, Homework Help, Discussion**