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?

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?

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

Please correct me if I am wrong.

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.

Please ignore my last post as I sent it 2-3 days ago and just appear
(due to post witheld by sci.physics.research moderator).

On Sun, 15 Jun 2008 19:01:13 +0000, rambotrout wrote:

Please correct me if I am wrong.

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.

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).