Re: capacitance formula

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Bert Hickman by way of Terry Fritz <teslalist-at-qwest-dot-net>" <bert.hickman-at-aquila-dot-net>

Dan and Pete,

What I said and what I meant to say were apparently two different 
things...  :^)

What I relay meant to say is that the voltage distribution across the a 
sandwich of dielectrics is distributed in the same fashion that the voltage 
is distributed across an equivalent chain of series connected 
capacitors.  The dielectric with the lowest relative permittivity (i.e., 
smallest capacitance) will face the greatest voltage stress - the voltage 
stress (volts/mil) will be inversely proportional to the relative 
permittivities of the dielectrics.

Best regards,

-- Bert --
-- 
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Tesla list wrote:
>Original poster: "by way of Terry Fritz <teslalist-at-qwest-dot-net>" 
><dhmccauley-at-spacecatlighting-dot-com>
>
>Theoretically yes, in reality no.  You can never assume capacitors in series
>will share voltage equally among them.
>In practice,  you would also use equalizing resistors across each capacitor
>to equalize the voltage across the capacitors.
>The Captain
>  > >I've been wondering what the capacitance between two conductive plates
>  > >separated by some distance that is filled with both a sheet of plastic
>  > >and some oil, where the plastic and oil have different dialectric
>  > >constants, and the thickness of each is different.
>  > >I've come to the conclusion that it will be the same as if it were two
>  > >capacitors in series, one purely the plastic, the other purely the oil,
>  > >each with their own individual thicknesses. Then use the series
>capacitance
>  > >formula C = 1/ (1/c1 + 1/c2).
>  > >Is this correct?
>  > >-Pete Lawrence.
>  > >
>  > >.
>  >
>  > Pete,
>  >
>  > Yes.  And the voltage across the individual dielectric systems will share
>  > identically as with capacitors connected in series.
>  >
>  > -- Bert --
>
>.