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Date: Mon, 28 Mar 2011 09:10:53 +0200
From: "Johan H. Bodin" <jh.bodin@telia.com>
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Subject: Re: LF: Earth capacitance
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Interesting thoughts...

I had a vague, and probably wrong, memory that the electron density of the
ionosphere had a maximum around 300km so I placed the outer sphere at that
altitude and calculated "only" 15.8mF. Still, the energy stored is quite
impressive. W = V^2*C/2 and V = 250kV => 494 MJ = 137 kWh, enough to heat my
house for a couple of days... Now, where do I put my crocodile clip to
"download" this energy? ;-)

73
Johan SM6LKM

----

Andy Talbot skrev 2011-03-27 23:42:
> 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 = 4.pi.6371000^2  m^2
> Height of ionosphere   d = 60000m
> 
> C = Eo . A / d    = 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 = Q/(4*pi*eps*R*R)
>> having now two spheres: inner (earth) R1 and outer: ionosphere R2
>> the potential V = integral from R1 to R2 over EdR which immediately gives
>> V = (Q/4*pi*eps)*(1/R1 - 1/R2) so as C = Q/V one gets
>> C  = 4*pi*eps/(1/R1 - 1/R2)
>> assuming R1 = 6370km ; R2 = 6370km + 60km = 6430 km; eps = 8.854*10^-12 F/m
>> C is about 0.076F or 76 miliFarad
>>
>>                              pse, check my "homework"  :)   Piotr, sq7mpj
>>
>> qth: Lodz /jo91rs/
>>
>>
>>
>>
> 
>