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 va***m case. The Coulumb constant is affected by
> electric constant (va***m 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
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