On Thu, 12 Jun 2008 14:40:01 -0700 (PDT), rambotrout
<rambotrout@[EMAIL PROTECTED]
> 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.
>
>> > 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?
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
My postings are going to some alt. group for some reason
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