Re: capacitance formula

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

Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

At 03:01 PM 6/4/2003 -0600, you wrote:
>Original poster: "Peter Lawrence by way of Terry Fritz 
><teslalist-at-qwest-dot-net>" <Peter.Lawrence-at-Sun.COM>
>
>
>One thing I am not claiming is that if the space between the conductive
>plates is filled with materials of two different dialectric constants
>that are not essentially flat (plastic sheet and oil are both essentially
>flat, but powered or granular material suspended in oil or glass fibers
>in an epoxy or polyester matrix for example are not) then the simple formula
>still holds, in fact it probably fails in this case. In this case the lines
>of the electrostatic field are not straight and parallel, and the two
>different dialectrics can no longer be (conceptually) separated into
>two independent capacitors.


Actually, this is the case.. you can use the "mixing" formulas to calculate 
the effective epsilon of a reasonably homogenous mixture.  Here's an 
intuitive argument as to why:

Consider first a stack of two layers of material with two dielectric 
constants.  The overall capacitance (per unit area) is epsilon0 * 
1/[1/(epsilon1/thickness1)+1/(epsilon2/thickness2)]
That is, the total capacitance is the series combination of the two 
component capacitances.

Now imagine splitting each of the two layers into two layers each.  The 
capacitance of each layer is proportional to (epsilon/thickness)... 
Essentially, it's like having two capacitors in series to replace one 
smaller capacitor.  You've now got a series combination of 4 capacitors, 
and the order in which you have them is immaterial.  It makes no difference 
if the arrangement of layers is {1a, 1b, 2a, 2b} or {1a, 2a,1b,2b}.

So, we can divvy up the space between the plates into as many tiny layers 
of various thicknesses as we like and you combine the epsilons according to 
1/thickness...

So, what's the effective epsilon of a stack of two...
epsilon/thickness = 1/(1/(epsilon1/thickness1) + 1/(epsilon2/thickness2))
or
thickness/epsilon = thickness1/epsilon1 + thickness2/epsilon2



Now, consider a capacitor with two different dielectrics arranged 
horizontally. This looks like two capacitors in parallel.  The capacitance 
is epsilon0 * (area1*epsilon1/thickness + area2*epsilon2/thickness).  The 
thickness is the same, everywhere, so the effective epsilon is just 
area1/(area1+area2)*epsilon1 + area2/(area1+area2)*epsilon2