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*To*: tesla-at-pupman-dot-com*Subject*: Re: Cap calculations*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sun, 25 Feb 2001 16:34:06 -0700*Resent-Date*: Sun, 25 Feb 2001 17:12:22 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <RSsDZD.A._1G.O9Zm6-at-poodle>*Resent-Sender*: tesla-request-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

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