Dear rambotrout:
On Jun 12, 2:40=A0pm, rambotrout <rambotr...@[EMAIL PROTECTED]
> wrote:
> > > What would be the electric field in between the
> > > dielectric materials?
>
> > No change, I think. =A0The electric field is impressed
> > by the charge on the plates. =A0The 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 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 =3D 80.
salt water, k =3D 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
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