Re: Cap calculations

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

Original poster: "Georg Sickinger by way of Terry Fritz <twftesla-at-uswest-dot-net>" <sickinger-at-web.de>

>
> Hello. I have a major problem. For one of my experiments I need to know the
> exact capacity of a cap consisting of two big allumininum plates with air
> between them.

It is:
C = 8,85 As /Vm * [Area in square meter]  / [spacing in meter]
This equation yields farads (As/V), that means, if you need mF or nF you have
to multiply the whole by the corresponding factor.
[8,85 As/Vm] is the dielectric constant of air.
If there is another material between your cap plates you will have to
muliply the
result with the dielektic factor of this material.

> The only known variables are the size of the plates and the distance between
> them which I want to transfer into farads.
> Furthermore I want to know the definition of one farad. Someone told me it is
> the capacity of two plates sized one square meter and placed at one meter
> distance. Is this true ?

Definition:
1 farad is the capacity which can save the charge of 1 Coulomb (= As) at a
tension of 1 volt.
If one puts your parameters into the equation (m ², m), the capacity of 8.85 pF
is the result !!.That means, if the plates have a separation of 1 m you need an
area of 113 000 square kilometers (!!!) to reach a farad.

> Oh,  and one more thing: please make your calculations in cm and mm because I
> am European and do not work with inches.

In consideration on the formulae and common unities I have used [m], but this
might not represent any problem : -)

> Sietze van de Burgt

Georg, Germany