Gene Fuller wrote:
... your long exposition above
is not what you said previously, E*H*sin(A).

I think you know that I meant the magnitudes only,
but to be entirely technically correct it should
have been:

Poynting vector = ExH = |E|*|H|*sin(A)

The same question remains: When E and H are both
not zero, how can the Poynting vector be zero?
--
73, Cecil http://www.w5dxp.com