# Quality Factor of Cavity Resonator

1. Mar 30, 2009

### maverick280857

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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?

2. Apr 1, 2009

### maverick280857

Solved>

3. Nov 30, 2010

but how?