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# Re: capacitor

```At 11:11 AM 01/02/2000 +0100, you wrote:
>I've got two 0,5 m^2 copper plates and window glass in between (0,4cm =
>4mm). How big is this cap?
>
>greets
>Tom

Hi Tom,

You have perhaps already found the answer but for all our listeners...

Assuming the copper plates are in direct contact with the glass without
many significant air bubbles...

The "k" value for window glass is 7.8 (that number varies a little bit
depending one who's it is).

The equation is:

C = 8.85x10^-12 x K x A / D

C = The capacitance in Farads
K = The dielectric constant which is dimensionless and works for all unit
systems (English, metric, mm, inches...)
A = overlapping plate area in square meters
D = Distance between plates in meters.
8.85 x 10^12  Is one of those basic constants of the universe called the
"permittivity of free space".  The dimension is Farads per Meter.

So,

C = 8.85e-12 x 7.8 x 0.5 / 0.004 === 8.63e-10 Farads or 0.86nF or 860 pF.
Not a very large capacitance.  You usually need about 10, 15, or more
plates like this for a TC.  The puncture voltage is 34000 volts so your
fine there.  How about a nice MMC!!

BTW - Hopefully, this will be one thing we can all agree on...  However, it
is always fun when someone asks a question like this to see all the
different answers "we geniuses" come up with! :-)))

The equation is from Steve Bell's web site at
"http://www.breakfix.demon.co.uk/tesla_cap1.html" and the constant is from
Mike Hammer's site at "http://www.monmouth-dot-com/~grimcorona/capacito.htm".
Hopefully, I can blame THEM if the above is wrong :-))))

Cheers,

Terry

```

References:
• Re: capacitor
• From: Parpp807-at-aol-dot-com (by way of Terry Fritz <twftesla-at-uswest-dot-net>)