If two electrodes are sandwitching two dielectric
materials with very different dielectric constants
(but the same thickness), say, water and glass.

Would the new dielectric constant lies in
between the original two?

What would be the electric field in between the
dielectric materials?

I suppose not half of the total electric field
imposed by the electrodes. Would the larger
dielectric constant material take up more
of it?

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

Is there any relation between dielectric
constant and dielectric strength?

What would be the electric field in between the
dielectric materials?

No change, I think. The electric field is impressed by the charge on
the plates. The amount of energy involved in impressing that
particular field, that is something else again.

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

No, it controls its "leakage" or resistivity.

What would be the electric field in between the
dielectric materials?

No change, I think.  The electric field is impressed
by the charge on the plates.  The amount of energy
involved in impressing that particular field, that is
something else again.

Do you mean it follows the Coulumb's law without
being affected by the dielectric material? I thought
(but I may be wrong) the dielectric material would
change the electric field in the material as the law
is derived for the vacumm case. The Coulumb
constant is affected by electric constant (vacumm
permittivity) and a dielectric onstant is the ratio of
static permittivity of the material and electric
constant. I am pretty sure it does change
something just like it affects the capacitance.

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

No, it controls its "leakage" or resistivity.

I don't think I am getting an answer. Assume
that the electrodes are thinly insulated so as
to block current leakage. Would water with ions
in it still retain its dielectric constant of 80?

If two electrodes are sandwitching two dielectric materials with very
different dielectric constants (but the same thickness), say, water
and glass. Would the new dielectric constant lies in between the
original two?
You can insert a 3d plate in the sandwich, then analyze 2 capacitors

What would be the electric field in between the dielectric materials?
I suppose not half of the total electric field imposed by the
electrodes. Would the larger dielectric constant material take up more
of it?

If the water contains ions, would that change its dielectric constant
from that of its pure form (about 80)?

Is there any relation between dielectric constant and dielectric
strength?
John Polasek

On Thu, 12 Jun 2008 11:53:10 -0700 (PDT), rambotrout

rambotr... at (no spam) yahoo.com> wrote:
If two electrodes are sandwitching two dielectric materials with very
different dielectric constants (but the same thickness), say, water
and glass. Would the new dielectric constant lies in between the
original two?

You can insert a 3d plate in the sandwich, then analyze 2 capacitors
in series.
The charge Q and displacement D = Q/A = Ei*Ki/A are constant
throughout including on the plates. The individual voltages are
inverse to the dielectric constant. Vi = Ei*Ti (T = thickness. A =
area).

What would be the electric field in between the
dielectric materials?

No change, I think. The electric field is impressed by the charge on
the plates. The amount of energy involved in impressing that
particular field, that is something else again.

Do you mean it follows the Coulumb's law without being affected by the
dielectric material? I thought (but I may be wrong) the dielectric
material would change the electric field in the material as the law is
derived for the vacumm case. The Coulumb constant is affected by
electric constant (vacumm permittivity) and a dielectric constant is
the ratio of static permittivity of the material and electric
constant. I am pretty sure it does change something just like it
affects the capacitance.

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

No, it controls its "leakage" or resistivity.

I don't think I am getting an answer. Assume that the electrodes are
thinly insulated so as to block current leakage. Would water with ions
in it still retain its dielectric constant of 80?

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?

What would be the electric field in between the
dielectric materials?

No change, I think. The electric field is impressed by the charge on
the plates. The amount of energy involved in impressing that
particular field, that is something else again.

Do you mean it follows the Coulumb's law without being affected by the
dielectric material? I thought (but I may be wrong) the dielectric
material would change the electric field in the material as the law is
derived for the vacumm case. The Coulumb constant is affected by
electric constant (vacumm permittivity) and a dielectric constant is
the ratio of static permittivity of the material and electric
constant. I am pretty sure it does change something just like it
affects the capacitance.

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

No, it controls its "leakage" or resistivity.

I don't think I am getting an answer. Assume that the electrodes are
thinly insulated so as to block current leakage. Would water with ions
in it still retain its dielectric constant of 80?

What would be the electric field in between the
dielectric materials?

No change, I think. The electric field is impressed by the charge on
the plates. The amount of energy involved in impressing that
particular field, that is something else again.

Do you mean it follows the Coulumb's law without being affected by the
dielectric material? I thought (but I may be wrong) the dielectric
material would change the electric field in the material as the law is
derived for the vacumm case. The Coulumb constant is affected by
electric constant (vacumm permittivity) and a dielectric constant is
the ratio of static permittivity of the material and electric
constant. I am pretty sure it does change something just like it
affects the capacitance.

If the water contains ions, would that change
its dielectric constant from that of its pure
form (about 80)?

No, it controls its "leakage" or resistivity.

I don't think I am getting an answer. Assume that the electrodes are
thinly insulated so as to block current leakage. Would water with ions
in it still retain its dielectric constant of 80?
You can insert a 3d plate in the sandwich, then analyze 2 capacitors

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?

Yes.  "Electrodeionization".

In this case, wouldn't the dielectric
constant of the water is retained?

Yes, as I gave you numbers before:
pure water, k = 80
salt water, k = 81

The water might break down under fairly weak
fields because K = 80 is a very high number for
a simple natural substance.

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.

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.

What is "thinly insulated"?

Dear rambotrout:

"rambotrout" <rambotrout at (no spam) yahoo.com> wrote in message
news:20ed4345-0da2-47d2-942c-ce21db760ae4 at (no spam) j22g2000hsf.googlegroups.com...
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?

Yes. "Electrodeionization".

In this case, wouldn't the dielectric
constant of the water is retained?

Yes, as I gave you numbers before:
pure water, k = 80
salt water, k = 81

David A. Smith
The water might break down under fairly weak fields because K = 80 is

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.

This sentence doesn't parse:
This is how energy is liberated from "charging" a
dielectric.

Please explain, preferably without the assistsance
of poultry.

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

will necessarily raise the field intensity elsewhere

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

Dear John C. Polasek:

On Jun 19, 6:48 am, John C. Polasek <jpola... at (no spam) cfl.rr.com> wrote:
...
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?

Yes.  "Electrodeionization".

In this case, wouldn't the dielectric
constant of the water is retained?

Yes, as I gave you numbers before:
pure water, k = 80
salt water, k = 81

The water might break down under fairly weak
fields because K = 80 is a very high number for
a simple natural substance.

They are two different measures. Water is polar, so without breaking
down it can rotate to where most of the oxygen atoms are
preferentially oriented towards the anode.

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

Roughly, imagine a bunch of frozen chickens hanging from "pegs" inside

an enclosure in empty space. The chickens will be randomly oriented.
Now move it into a gravity field, and the chickens will hang "down".
In this (silly) analog, the center of mass is slightly lower in the
graivtational field... because the average chicken is closer to the
"bottom".
This sentence doesn't parse:
This is how energy is liberated from "charging" a
dielectric.
Please explain, preferably without the assistsance of poultry.
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.

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

David A. Smith