- #1

- 22

- 0

Thank you,

Chet

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter littlebilly91
- Start date

- #1

- 22

- 0

Thank you,

Chet

- #2

- 5,570

- 200

Ok, I am looking over stuff for an exam. I am a bit confused on something. When conductivity is zero, a medium is lossless.Correct

Zero conductivity means infinite resistance. For infinite resistance, no electrons will flow, but an electric field can propagate forever.Correct

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,No, when conductivity is zero, attenuation constant is zero. though it is very small for small frequencies. Is everything here correct or am I overlooking something?

Thank you,

Chet

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

- 22

- 0

- #4

- 5,570

- 200

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

[tex]\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}[/tex].

Where the solution is a pair of phasor equations.

[tex]\tilde E(z)=Ae^{\delta z}+Be^{-\delta z}[/tex]

For good conductor:

[tex]\sigma=\infty\;\Rightarrow \alpha=\beta=\sqrt{\pi f \mu\sigma}[/tex]

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.

[tex]\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}[/tex].

Where the solution is a pair of phasor equations.

[tex]\tilde E(z)=Ae^{\delta z}+Be^{-\delta z}[/tex]

For good conductor:

[tex]\sigma=\infty\;\Rightarrow \alpha=\beta=\sqrt{\pi f \mu\sigma}[/tex]

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:

- #5

NUCENG

Science Advisor

- 914

- 0

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

- 5,570

- 200

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.

Oh yeh, I remember now. It is from early chapters in EM book. For good conductor, you cannot have

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

To the OP, this is the simple answer.

Last edited:

- #7

- 22

- 0

Share:

- Replies
- 10

- Views
- 1K