Dear John C. Polasek:
On Jun 19, 1:34=A0pm, John C. Polasek <jpola...@[EMAIL PROTECTED]
> wrote:
> On Thu, 19 Jun 2008 07:30:59 -0700 (PDT),dlzc<dl...@[EMAIL PROTECTED]
> wrote:
> >On Jun 19, 6:48=A0am, John C. Polasek <jpola...@[EMAIL PROTECTED]
> wrote:
=2E..
> >> What this means is that the bound electrons
> >> are on very weak "springs" and have large
> >> deflections (K =3D 80) so that a moderate field might
> >> break the springs and free the electron for
> >> conduction.
>
> >No. =A0You are conflating conduction or conductivity
> >with permittivity. In conduction (your "breakdown"),
> >electrons / ions are free to migrate through the
> >material the electric field is applied to. =A0In
> >permittivity, the material undergos a "physical"
> >change NOT requiring the motion of loose charges.
>
> You seem unable to read a sentence: I am saying
> you have permittivity as long as the electrons
> remain elastically bound, (and thus able to
> store energy) but upon their breaking loose you
> have the ohmic condition.
I can read a sentence. You veered off from permittivity to discussing
breakdown. This is not what the OP asked about.
Additionally, the Rube Goldberg device you construct, relating high k
value (loose springs) to low dielectric breakdown voltage does not pan
out.
=2E..
> This sentence doesn't parse:
> >This is how energy is liberated from "charging" a
> >dielectric.
>
> Please explain, preferably without the assistsance
> of poultry.
In general, the water molecules (in this case) do not get closer
together, they simply orient themsleves with the oxygen atoms facing
the anode. The analogy you ceased to be humored by used gravitation
in place of an applied E field. In a material, alignment of charges
yields energy... like the "latent heat of fusion" of a salt, for
example.
=2E..
> >> If the battery remains connected, the E field is still
> >> there, exacerbated by local ionizations, and able
> >> to do more ionization rather than herding ions where
> >> they can't do any harm.
>
> >Mobile ions / electrons serve to reduce the field in
> >the dielectric, since they tend to hover very near
> > the plate.
>
> Any incidence of shortening the gap, as I assume
> could occur with ions
Not really, no. They really will "plate out" on the electrode about
as close as possible.
> will necessarily raise the field intensity elsewhere
Absolutely not. The ions themselves have opposite sign, and *directly
reduce* the E-field. The are not polar, that have a single (sometimes
more) unbalanced charge.
> with possible avalanche results.
>
> >> What is "thinly insulated"?
>
> >For a capacitive cell, with a dielectric thickness
> >t_d, and "thin insulator" thickness t_i:
> >2 * t_i << t_d
So at least in your opinion I answered that one...
David A. Smith
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