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?

Dear rambotrout:

On Jun 12, 11:53*am, rambotrout wrote:
If two electrodes are sandwitching two dielectric
materials with very different dielectric constants
(but the same thickness), say, water and glass.

... better, air and glass.

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

Yes. Over the total separation.

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.

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

The dielectric controls the current that will flow for a given applied
voltage.

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.

Is there any relation between dielectric
constant and dielectric strength?

Not really, or at least not directly.

http://www.ami.ac.uk/courses/topics/0184_dp/index.html
http://en.wikipedia.org/wiki/Dielectric_strength
http://en.wikipedia.org/wiki/Dielectric_constant

Dielectric strength has to do with the strength of the weakest bond.
Dielectric constant has to do with how polar an atom or molecule is.

David A. Smith

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?

Dear rambotrout:

On Jun 12, 2:40*pm, rambotrout wrote:
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.

The electric field is governed by the charge on the plates. I had
assumed you left a "battery " connected, and were interested only in
how the "electric field" was distributed within the medium.

Maybe you need to wait on a better answer on this one from someone
lese.

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?

http://lists.contesting.com/_topband.../msg00111.html
fresh water, k = 80.
salt water, k = 81.

The k value describes how the material stores energy under an electric
field. The water molecule "deforms", as well as aligning. Ions will
only align.

David A. Smith

On Jun 13, 2:37 am, John C. Polasek wrote:
On Thu, 12 Jun 2008 11:53:10 -0700 (PDT), rambotrout

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 are Ki and Ei? How do you get displace D = Q/A?

On Thu, 12 Jun 2008, rambotrout wrote:

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.

For Coulomb's law in a dielectric medium, the D-field (i.e., the electric
displacement) is unaffected by the medium (recall that the relevant
Maxwell equation is div(D) = rho, where rho is the free charge density).
Since D = eE, e = permittivity, E is reduced as e increases. The electric
force is F = qE, so the force is reduced. Just use e instead of e0 in
Coulomb's law (Coulomb's constant being 1/(4*pi*e0)).

As your your original question, what stays constant? The charge on the
plates? Or the voltage across them? If Q is constant, then D will remain
the same, and it'll be easy to find E everywhere between the plates. When
you know E, you can find the potential V at any point easily.

Since we know that the field between two parallel plates in free space
(ignoring edge effects) is E = (Q/A)/e0, we have D=Q/A.

For V held constant, then you have V = E1*d1 + E2*d2, where E1 and E2 are
the fields within the dielectrics, and d1 and d2 are the thicknesses. E1
and E2 are both unknown, but the continuity of D (basically, D1=D2, which
gives e1*E1 = e2*E2) gives the required extra equation. This will work for
as many layers as you care to include.

If you want an "average"/"effective" dielectric constant, how do you want
to define it? The "average" E can be taken to be E_eff = V/distance =
V/(d1+d2), D is constant, so D = e_eff E_eff looks good.

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?

Water with ions is conductive. If there are enough ions, you can treat it
as a perfect conductor - the conductivity will be high enough so the
charge distribution in the water will reach equilibrium. What is the
dielectric constant of a perfect conductor?

If the number of ions is small enough, the conductivity will be low enough
so that it can be ignored for reasonable times. If the ions don't move
significantly, don't expect any major effect on the dielectric constant.
Somewhere in between would be the difficult case where you can't treat the
water+field as an electrostatic problem (unlike both the perfect conductor
and insulator limits) since equilibrium would not be reached during the
times of interest.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/...,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

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?

Dear rambotrout:

"rambotrout" wrote in message
...
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 befo
pure water, k = 80
salt water, k = 81

David A. Smith

Dear John C. Polasek:

On Jun 19, 6:48*am, John C. Polasek 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 befo
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.

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 is how energy is liberated from "charging" a
dielectric.

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

Dear John C. Polasek:

On Jun 19, 1:34*pm, John C. Polasek wrote:
On Thu, 19 Jun 2008 07:30:59 -0700 wrote:
On Jun 19, 6:48*am, John C. Polasek 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

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

David A. Smith