Dielectric
rambotrout
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?
dlzc
On Jun 12, 11:53 am, rambotrout <rambotr...@yahoo.com> 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
rambotrout
> > 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?
dlzc
On Jun 12, 2:40 pm, rambotrout <rambotr...@yahoo.com> 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/2002-07/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
John C. Polasek
<rambo...@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?
>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
rambotrout
> On Thu, 12 Jun 2008 11:53:10 -0700 (PDT), rambotrout
>
> >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?
rambotrout
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?
Uncle Al
>
> 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)?
Water with ions is electrically conductive. A better example would be
two solid slabs, perhaps contrasting polyethylene foam (about 1.3,
coax cable) and poly(vinylidene fluoride) at 12.2 or potassium
tantalate niobate at 6000.
What if you insulated your DC electrodes with a couple of microns
thickness of Parylene-C film then dipped them in electrolyte solution
or placed a copper slab in-between?
> Is there any relation between dielectric constant
electric field attenuation
> and dielectric
> strength?
breakthrough voltage/thickness
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
Timo A. Nieminen
>>> 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/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
rambotrout
Say the voltage across the slabs is Vt, then Vt = V1 + V2, where V1,
V2 are voltages across the two different dielectric materials.
Therefore,
Vt = Q*d1/(E1 * A) + Q*d2/(E2 * A),
where Q = charge in Coulomb, d = thickness of material, E = dielectric
constant, and A = area.
rambotrout
(due to post witheld by sci.physics.research moderator).
rambotrout
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?
N:dlzc D:aol T:com (dlzc)
"rambotrout" <rambo...@yahoo.com> wrote in message
news:20ed4345-0da2-47d2...@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
John C. Polasek
You can insert a 3d plate in the sandwich, then analyze 2 capacitors
in series. Let the factors by K1, T1 and K2, T2.
C1 = K1/T2 C2 = K2/T2
C2/C1 = K2T1/K1T2 = V1/V2 (volts inverse to cap. for same charge)
This gives you the voltage split. You can work from that.
John Polasek
rambotrout
about the relative permitivity of salt water, David. Really appreciate
all helps.
Andrzej Novak
Although it may not matter depending on the situation you are considering
(especially for DC circuits), but you might want to consider searching
for "complex dielectric constant"---this especially matters in AC
circuits as the nonzero conductivity could lead to dissipative losses.
Circuit-wise, look at it as this: two capacitors (water and the thin
insulator) connected in series, and a small resistor connects two ends of
one capacitor (the one with water). In a DC circuit, you can ignore the
resistor, but not if you have AC voltage source.
Andrzej Novak
You are on the right track. In order to calculate the equivalent
dielectric constant for the whole assembly, you can treat the system as
two capacitors in series, each with a single kind of dielectric. I
usually imagine an infinitesimally think conductor between them.
Remember that capacitors in series add in reciprocals (1/C_eq = 1/C_1
+ 1/ C_2), and you can work out what the effective dielectric constant
is (which would be relatively simple if d1 == d2, by the way).