# Conduction and lossless materials

1. Dec 11, 2011

### littlebilly91

Ok, I am looking over stuff for an exam. I am a bit confused on something. When conductivity is zero, a medium is lossless. Zero conductivity means infinite resistance. For infinite resistance, no electrons will flow, but an electric field can propagate forever. A wire has high conductivity, so electrons can flow easily. I understand the difference between electrons and electric fields, but this seems counter-intuitive. Also, when conductivity is zero, the attenuation constant is not zero, though it is very small for small frequencies. Is everything here correct or am I overlooking something?
Thank you,
Chet

2. Dec 11, 2011

### yungman

Because good conductor like metal can conduct current, conductivity is high. loss is very high, therefore attenuation constant is very high and EM wave get attenuated when going into the good conductor.

There is almost no material that is truely lossless, just matter how low the loss. If there is a true lossless material, attenuation constant is zero. Good luck in finding one other than vacuum.............well air maybe be consider lossless too!!!

3. Dec 11, 2011

### littlebilly91

Ok, cool. The only issue I have is why does conductive material have high loss? When electrons can easily flow, I would expect loss to be low.

4. Dec 11, 2011

### yungman

Explanation is actually not easy, it is all electromagnetics theory of EM wave in various medium where charge free wave equation:

$$\nabla^2\tilde E -\delta^2\tilde E=0\;\hbox { where } \;\delta=\alpha +j\beta = j\omega\sqrt{\mu \epsilon_0\epsilon_r}\;\sqrt{1-j\frac{\sigma}{\omega\epsilon_0\epsilon_r}}\;\hbox { which } \alpha \;\hbox { is attenuation constant and }\;\beta \;\hbox { is the phase constant}$$.

Where the solution is a pair of phasor equations.

$$\tilde E(z)=Ae^{\delta z}+Be^{-\delta z}$$

For good conductor:

$$\sigma=\infty\;\Rightarrow \alpha=\beta=\sqrt{\pi f \mu\sigma}$$

I skip a lot of steps to arrive to this, if you are interested, I can post the derivation in detail. But from this, you can see as conductance goes up, attenuation constant goes up so EM wave disappeared quickly when penetrating into a good conductor. Also, the velocity of the EM propagation slow down quite a bit as conductance goes up.

I still yet to find a lot of the EM theory intuitive!!!! If someone has a more common sense explanation, I love to learn myself.

Last edited: Dec 11, 2011
5. Dec 11, 2011

### NUCENG

The simple explanation I used to pass the tests in school is that the loss here is in terms of field strength. You may be confusing this with resistive losses from current flow.

Perfect conductor - zero field and high field loss, but no energy loss

Perfect insulator - infinite field and low field loss, but no energy loss

Real conductors and RLC components - somewhere in between with energy loss from heating.

It may not be perfectly accurate, but I've never worked with a perfect conductor or insulator since school.

6. Dec 11, 2011

### yungman

Oh yeh, I remember now. It is from early chapters in EM book. For good conductor, you cannot have E inside as any E developed inside a good conductor will cause electrons to move in direction so it neutralize the field. So the E disappeared real fast in good conductor and therefore the attenuation constant is very high.

I never link this together till now. Thanks for reminding me this.

To the OP, this is the simple answer. E decrease very fast in good conductors because electrons can move freely in good conductor. Any E developed inside a good conductor will immediately cause electrons to move in direction that will neutralize the E field. So the E will greatly attenuated fast. For perfect conductor, no E field can even develop inside and therefore the skin depth is zero and all current are on the surface.

Last edited: Dec 11, 2011
7. Dec 11, 2011

### littlebilly91

Ahhh... That makes sense. In lossless media, no electrons move, so the E field never looses energy. E&M is so cool. Thank you both for your responses!