Re: Cap calculations

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

Original poster: "Andreas Marshall by way of Terry Fritz <twftesla-at-uswest-dot-net>" <marand-at-t-online.de>

Capacitance in this case  is defined as

C=Epsilon(0)*Epsilon(r)*A/d

where

Epsilon(r) is the relative dielectric constant of the used material between
the plates.
Epsilon(0) is a constant with a value of: 8.85*10-12F/m.
A is the area of the plates in mē.
d is the distance of the plate in m.

This means:

A capacitor formed of 2 plates of 1mē in a distance of 1m in air has a
capacitance of:

8.85*10-12F/m*1*1mē*1m=8.85pF.

In a Distance of 1cm, this means 0,885nF.




Tesla list wrote:

> Original poster: "s. van de burgt by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <s.van.de.burgt-at-hccnet.nl>
>
> 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.
> 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 ?
> Oh,  and one more thing: please make your calculations in cm and mm because I
> am European and do not work with inches.
>
> Sietze van de Burgt