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Violation of Gauss Law

时间:04-04 整理:3721RD 点击:
Consider a horizontally propagating Gaussian beam of EM waves of vertical polarization.
Consider a box with rectangular horizontal top and bottom,
and with side walls going in the direction of the electric field E.
Let the box be above the axis of the Gaussian beam.
If the box is tall enough, then the magnitude of flux of E through the bottom
of the box will be much greater than the magnitude of flux through the top.
Since side walls go in the direction of the E field, there will be
zero flux through each of them.

I assume the box is empty.

I agree that the electric flux density will be different, but will the total flux? And what is the source of the requirement that the E-fields are always exactly aligned with the side walls?

Flux density will quickly decrease as we go up from the axis of the beam.
The side walls should be aligned with the lines of the E field so that the flux
through each of them is zero. Thanks to this the total flux is the difference
between the flux through the bottom and through the top.

Then either the box is not rectangular, or the beam is not gaussian. The E-field in a vacuum is only completely transverse to a single rectangular axis for a plane wave. I can't recall, or envision, this being the case for a gaussian beam.

Polarization of the beam is vertical, so lines of E field will be in vertical planes.

Its not a law, so much as a GUIDELINE.


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