Josephson and Howie

10.52am   23.8.2020

I should also say, since you people are so keen on niggling, that to be strictly correct, after the capacitor has been charged there will be transients that will decay rapidly, so in that sense, if you wish to interpret ’static’ in that way, there will never be a completely static field.  And again, thermal fluctuations and lightning strikes will lead to fluctuations.  Will that satisfy y’guys who insist there is no such thing as a static field in a capacitor?

- Brian Josephson.

This is very interesting. Both Josephson and Howie (wrongly) think of “ohmic losses” causing the energy, which they presumably agree enters the charging capacitor http://www.ivorcatt.org/icrwiworld78dec1.htm at the speed of light, to slow down. Let us enter their world, assuming they are right in this.

This means that (for them) in a Platonic, or ideal situation, of perfectly conducting capacitor plates, the theoretical perfect charged capacitor would permanently have a reciprocating, not a static, electric field. Thus, in their view, it is the failure of the capacitor to have perfectly conducting plates that leads to the static electric field.

This is not delivered as a reduction ad absurdum. It is delivered in the spirit of science. Dialogue is important.

As a further thought, I have difficulty linking resistance, and loss, with drop in velocity. When energy travelling at c meets resistance, does it not only diminish (some of it turning into heat or light) but also slow down? Surely all the ExH energy delivered into charging the capacitor would turn into heat or light, leaving the capacitor electrically uncharged?

The formula for the total of ExH energy in a reciprocating charged capacitor is equal to the formula for a capacitor with a static electric field.  The formula for the ExH energy entering the capacitor (mathematically) equals the formula for the energy in a charged capacitor. There is no loss of energy (if and) when the energy entering at velocity c slows down to static. (Wrong. During charging, the practical capacitor heats up a little, as it does when discharging. Similarly in Wakefield 4, where reciprocating energy turns into heat {= sinusoidal energy} which radiates out of the capacitor.)

Ivor Catt  12.00  23.8.2020

Afterthought.

A TEM wave travelling down between two resistive conductors gradually slows down. As some of the energy is dissipated in the conductors (or perhaps the dielectric) the remining ExH energy is slowing towards a standstill.    12.30pm same day.