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 <[email protected]> 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/
>
>
>
>
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