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.