From: Rik Strobbe <Rik.Strobbe@fys.kuleuven.be>
To: "rsgb_lf_group@blacksheep.org" <rsgb_lf_group@blacksheep.org>
Date: Mon, 28 Mar 2011 00:02:35 +0200
Thread-Topic: LF: Earth capacitance
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Hello Andy, Piotr,

two ways, one result ;-)
I would say Andy did it the engineers way and Piotr did it the scientists=
 way.
I hope I haven't insulted one of you (or both).

73, Rik  ON7YD - OR7T
________________________________________
Van: owner-rsgb_lf_group@blacksheep.org [owner-rsgb_lf_group@blacksheep.or=
g] namens Andy Talbot [andy.g4jnt@gmail.com]
Verzonden: zondag 27 maart 2011 23:42
Aan: rsgb_lf_group@blacksheep.org
Onderwerp: Re: LF: Earth capacitance

A slightly different way of doing it  - from EM first principle,  but
yes spot on
Very close to both my calc and the quoted figure seen in a paper.

Worked out from the parallel plate capacitor equation :
Area of Earth surface =3D 4.pi.6371000^2  m^2
Height of ionosphere   d =3D 60000m

C =3D Eo . A / d    =3D 0.075F.
Which is a large electrolytic as seen in a typical big amateur 12V linear=
 PSU

And seemed awfully high when I first saw the figure.   Especially when
charged to quarter of a MV

Andy
'JNT



On 27 March 2011 22:18, Piotr Mlynarski <sq7mpj@toya.net.pl> wrote:
> Andy Talbot pisze:
>>
>> Here's an interesting little thought excercise
>>
>> Consider the Earth as one plate of a capacitor, and the lower surface
>> of the ionosphere as the other.   Without actually doing the
>> calculation, what do you 'feel' the capacitiance might be?
>>
>> Then do the calculation; the result is rather surprising
>>
>>
>
> Dear Andy, LF
>
> as i had completely no idea what might be the value of such defined
> capacitance
> i immediately thought that this "interesting little thought excercise"
>  could be added to
> my sunday evening cup of coffee ... :)
>
> Gauss theorem says that E field from charge Q on a sphere of radius R is
> E =3D Q/(4*pi*eps*R*R)
> having now two spheres: inner (earth) R1 and outer: ionosphere R2
> the potential V =3D integral from R1 to R2 over EdR which immediately gi=
ves
> V =3D (Q/4*pi*eps)*(1/R1 - 1/R2) so as C =3D Q/V one gets
> C  =3D 4*pi*eps/(1/R1 - 1/R2)
> assuming R1 =3D 6370km ; R2 =3D 6370km + 60km =3D 6430 km; eps =3D 8.854=
*10^-12 F/m
> C is about 0.076F or 76 miliFarad
>
>                              pse, check my "homework"  :)   Piotr, sq7mp=
j
>
> qth: Lodz /jo91rs/
>
>
>
>