Quality Factor of Cavity Resonator

[maverick280857]

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

$$Q = \frac{1}{\tan\delta}$$

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

Homework Equations

The Attempt at a Solution

$$Q = \omega\frac{W}{P_{loss}}$$

I have no idea what to do here, but I don't understand why textbooks attribute $P_{loss}$ 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.

 

[maverick280857]

Solved>

 

[mahinda]

but how?