This question asks about conducting cavity driven as an electromagnetic resonator.

I have never heard of conducting cavity before, so I was not familiar with that. Well now I know the boundary condition for a conducting cavity. But can someone give a physical explanation of what a conducting cavity is and how do we get the boundary conditions?

Also, what is an electromagnetic resonator?

Thanks.

## GRE9277 #34

### Re: GRE9277 #34

I realize that this is an old post, but I will post in the hope it helps someone else who is studying.

In terms of a conducting cavity, you can just think about any cavity where the surrounding walls are composed entirely of a perfect conductor.

The boundary conditions for a perfect conductor are as follows

$$$\vec{E^{\|}_2} - \vec{E^{\|}_1} = 0$\\[\tex]

$\vec{D^{\perp}_2} - \vec{D^{\perp}_1} = \sigma$$$

$$$\vec{B^{\perp}_2} - \vec{B^{\perp}_1} = 0$$$

$$$\vec{H^{\|}_2} - \vec{H^{\|}_1} = \vec{k} \times \vec{n}$ \hspace{1cm} where $\vec{k}$$$ is electric current density and $\vec{n}$ is a vector normal to the surface

For more information about electromagnetic resonators you may want to see this page:

https://en.wikipedia.org/wiki/Resonator

Please let me know if anyone has any questions.

Thanks.

In terms of a conducting cavity, you can just think about any cavity where the surrounding walls are composed entirely of a perfect conductor.

The boundary conditions for a perfect conductor are as follows

$$$\vec{E^{\|}_2} - \vec{E^{\|}_1} = 0$\\[\tex]

$\vec{D^{\perp}_2} - \vec{D^{\perp}_1} = \sigma$$$

$$$\vec{B^{\perp}_2} - \vec{B^{\perp}_1} = 0$$$

$$$\vec{H^{\|}_2} - \vec{H^{\|}_1} = \vec{k} \times \vec{n}$ \hspace{1cm} where $\vec{k}$$$ is electric current density and $\vec{n}$ is a vector normal to the surface

For more information about electromagnetic resonators you may want to see this page:

https://en.wikipedia.org/wiki/Resonator

Please let me know if anyone has any questions.

Thanks.