Dear John C. Polasek:

On Jun 19, 1:34 pm, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
On Thu, 19 Jun 2008 07:30:59 -0700 (PDT),dlzc<dl... at (no spam) cox.net> wrote:
On Jun 19, 6:48 am, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
...
What this means is that the bound electrons
are on very weak "springs" and have large
deflections (K = 80) so that a moderate field might
break the springs and free the electron for
conduction.

No.  You 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.  In
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.

...
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.

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
That blows away my conception of ideal geometry because,

So at least in your opinion I answered that one...
No it just raised the last question; does it need 2 empty spacers?
David A. Smith
John Polasek

On Thu, 19 Jun 2008 14:18:52 -0700 (PDT),dlzc<dl... at (no spam) cox.net> wrote:
Dear John C. Polasek:

On Jun 19, 1:34 pm, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
On Thu, 19 Jun 2008 07:30:59 -0700 (PDT),dlzc<dl... at (no spam) cox.net> wrote:
On Jun 19, 6:48 am, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
...
What this means is that the bound electrons
are on very weak "springs" and have large
deflections (K = 80) so that a moderate field might
break the springs and free the electron for
conduction.

No.  You 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.  In
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.

To the contrary, the OP asked asked this in
his original note: "Is there any relation between
dielectric constant and dielectric strength?"  

Again, failure to read.

And again I reiterate that high K means large
deflections with moderate fields such that
avalanching is possible as the bonds exceed
their elastic limit. That's how all breakdown occurs.

Additionally, the Rube Goldberg device you
construct, relating high k value (loose springs)
to low dielectric breakdown voltage does not
pan out.

I see you didn't consult my paper and I
consider it churlish of you to cut out reference
to it as the Permittivity paper at
http://www.dualspace.net.
Even a cursory examination will make it clear
that my model far outclasses your frozen
chicken hypothesis.

In any case, it's moot, all this talk of physical
chemistry and bonds, because the last
dielectric anyone would suggest would be
WATER! Water? Are you kidding?

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.

Yes, polarization stores energy and can also
release it, given a closed circuit.

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

That blows away my conception of ideal geometry
because, Wait, wait, I think you're trying to tell
me the dielectric needs 2 miniscule physical
separators to prevent the dielectric from touching
the electrodes. Would you mind amplifying on
that one?

So at least in your opinion I answered that one...

No it just raised the last question; does it need
2 empty spacers?

I would like to learn more about this "thin insulator"
concept. Do you have a reference text that
introduces it? It's not in Smyth or Panofsky.

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

Dear John C. Polasek:

On Jun 19, 5:19 pm, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
On Thu, 19 Jun 2008 14:18:52 -0700 (PDT),dlzc<dl... at (no spam) cox.net> wrote:
Dear John C. Polasek:

On Jun 19, 1:34 pm, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
On Thu, 19 Jun 2008 07:30:59 -0700 (PDT),dlzc<dl... at (no spam) cox.net> wrote:
On Jun 19, 6:48 am, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
...
What this means is that the bound electrons
are on very weak "springs" and have large
deflections (K = 80) so that a moderate field might
break the springs and free the electron for
conduction.

No.  You 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.  In
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.

To the contrary, the OP asked asked this in
his original note: "Is there any relation between
dielectric constant and dielectric strength?"  

I stand corrected.

Again, failure to read.

Ahem...

And again I reiterate that high K means large
deflections with moderate fields such that
avalanching is possible as the bonds exceed
their elastic limit. That's how all breakdown occurs.

But high K does not correlate with a low breakdown voltage.

Additionally, the Rube Goldberg device you
construct, relating high k value (loose springs)
to low dielectric breakdown voltage does not
pan out.

I see you didn't consult my paper and I
consider it churlish of you to cut out reference
to it as the Permittivity paper at
http://www.dualspace.net.
Even a cursory examination will make it clear
that my model far outclasses your frozen
chicken hypothesis.

Everyone has their opinion. And it was not an hypothesis, it was a
simile.

...
In any case, it's moot, all this talk of physical
chemistry and bonds, because the last
dielectric anyone would suggest would be
WATER! Water? Are you kidding?

I intentionally dropped an operating electric hand drill into a pool
of 18 meg ohm water. I then reached in and pulled the operating drill
out with my bare hand. No shock.

Additionally, some of the most used level detectors, use only water's
k value. "capacitive proximity detector" or "capacitive level switch"

Your personal experience limits your "common sense".

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.

Yes, polarization stores energy and can also
release it, given a closed circuit.

No poultry.

...
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

That blows away my conception of ideal geometry
because, Wait, wait, I think you're trying to tell
me the dielectric needs 2 miniscule physical
separators to prevent the dielectric from touching
the electrodes. Would you mind amplifying on
that one?

The intent is to have a net charge on each plate, just for "conceptual
purity".

So at least in your opinion I answered that one...

No it just raised the last question; does it need
2 empty spacers?

For the "mathematics", imagine that the anode is not insulated. Now
you will have one plate with a low positive charge and a high negative
charge. The dynamics changes, is all. Just a little weird to
consider.

I would like to learn more about this "thin insulator"
concept. Do you have a reference text that
introduces it? It's not in Smyth or Panofsky.

No, it is most likely either a pedantic device, or something that
sensor manufacturer's depend on to protect the conductors of their
electrodes. More than likely, the self-healing oxide layer of the
conductor when only low currents are applied.

The intent is to get at the field in the dielectric(s), and ignore the
ions which are not of concern (but can cause physical problems).

David A. Smith
Thank you. I recall a liquid level in the A1BNC system which was part

Thank you. I recall a liquid level in the A1BNC system
which was part of a bridge at 400hz.
In any real capacitor the function of the plates is to
contact every bit of the dielectric surface, simply as
a means of polarizing the dielectric.

An insulating film to protect against corrosion
makes sense but is in no way a factor in the
capacitors function. In that case it should be
included in the dielectric stack where because
of its near-micron thickness it could be ignored.

Electrolytic capacitors have high capacity this
way: the dielectric is chemically produced.
Wiki says the aluminum oxide can withstand
1e9 V/m which means in a 100 V capacitor
the film would be 1/10 micron!

In any case, it's moot, all this talk of physical
chemistry and bonds, because the last dielectric
anyone would suggest would be WATER!
Water? Are you kidding?

Dear John C. Polasek:

On Jun 19, 5:19 pm, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
...
In any case, it's moot, all this talk of physical
chemistry and bonds, because the last dielectric
anyone would suggest would be WATER!
Water? Are you kidding?

I was sitting here looking at the picture I have stored on my desktop,
and realized I could give you an example for you to consider:
http://www.sandia.gov/news/resources/releases/2006/physics-astron/hottest-z-output.html

You can get a really high resolution image of that picture (and it
makes a terrible desktop image... really hard to find icons in).

That "discharge array" occurs in deionized water. It does not form
single bolts from point to point, but is forced to spread out. Ask
your "common sense" how that could be.

David A. Smith
Courtesy response: Oh wow.

Thank you everyone for all the replies.

I have better grasp of it now.

Timo, if the electrode is thinly insulated, wouldn't all the ions get
attracted very close to their respective electrodes thus leaving the
water "relatively" pure? In this case, wouldn't the dielectric constant
of the water is retained?