Question $12$

$3$ pts

Waveguide cavities are commonly constructed using aluminum $(=3\mathrm{.}54{10}^7(\frac{S}{m})).$ In principle, a higher cavity Q factor could be achieved using a metal with higher conductivity such as silver $).$ Determine how much higher the Q would be if an aluminum cavity were replaced by a silver cavity. Assume all the dimensions $($and hence the resonant frequency$)$ are identical for both cavities. Express your answer as the ratio of the Q of the silver cavity to the Q of the aluminum cavity to two places after the decimal.

Hint: examine the equation for Q and determine how it varies with conductivity.