HI
why conductors are referred to as high loss mediums and dielectrics as low loss mediums?
what is getting lost when the medium is a conductor?

 

If the EM wave encounters a medium that is conductive, there are two parts that describe permittivity: the permittivity of the dielectric and the conductivity divided by omega times j (imaginary part).

If you solve the wave equation with the complex permittivity, you will see an exponential decay term that contains the imaginary part (conductivity / omega) and the real part permittivity. We can see, that the larger the imaginary part, the faster the component of the waveform will be decayed.

Steve

 

Wow Steve, it's Sunday (well here anyway), that's awfuly heavy stuff for a Sunday.

A simpler answer is that energy or power is being lost. The lost power is disappated as heat energy in the conductor by the 'conduction current' component in Steve's answer. The imaginary part of the solution produces what is known as 'displacement current' in the dielectric and represents stored energy or power.

Post again if you would like the maths.

 

Sorry, It's sunday here too, but I have been studying for about a week straight heavily and I guess it is showing.

It would be really nice to have an easy way to enter in equations in these posts, is that a possibility?? I would have done the math to show what I was trying to say with much more clarity.

Steve

 

It would be really nice to have an easy way to enter in equations in these posts

Hi Steve, I wrestled with this question as well, Dave was more than a bit helpful as this forum is set up to do just that.

Use the Tex options by clicking on the \(\Sigma\) in the advanced editing box.

I'm still learning about LaTex, but I understand it is currently all the rage in Academe.

I have also discovered that if you prepare the equation using the equation editor in Microsoft Word it pastes straight into this forum.