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
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