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Quality Factor of Cavity Resonator

1. Homework Statement

Prove that the quality factor of a rectangular cavity resonator which is filled with a lossy dielectric but has perfectly conducting walls, is given by

[tex]Q = \frac{1}{\tan\delta}[/tex]

where [itex]\tan\delta[/itex] is the loss tangent, i.e. the ratio of the imaginary part of the dielectric constant to the real part of the dielectric constant.

2. Homework Equations

3. The Attempt at a Solution

[tex]Q = \omega\frac{W}{P_{loss}}[/tex]

I have no idea what to do here, but I don't understand why textbooks attribute [itex]P_{loss}[/itex] only to the conducting wall loss. Or at least I haven't encountered a textbook where the dielectric loss -- not just the attenuation due to a lossy dielectric -- is calculated.

Also, it seems that this expression is true for both TE and TM Modes?

Please help.

Thanks in advance.
 
Solved>
 
but how?
 

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