pager48
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according to http://230nsc1.phy-astr.gsu.edu/hbase/electric/capsph.html#c2 isolated spheres have capacitance but how can it be measured with an instrument?
Very good point. An isolated sphere has a well defined capacitance but to measure anything one would have to connect an instrument and that means the sphere would no longer be isolated.pager48 said:isolated spheres have capacitance but how can it be measured with an instr
lesaid said:In my suggestion, when charging the spheres (while they are well isolated), there will be a very tiny capacitance from the (thin) wire that connects the sphere to the charging voltage, and that must have some small effect.
Then it does not matter if the sphere is hollow or solid to be a capacitor?lesaid said:No - it is the air around the sphere, the moisture in that air, anything other insulator near the sphere will contribute to the dialectric.
I believe that, for example, humid air around the sphere would result in a slightly higher capacitance than dry air, Though I stand to be corrected in that if any experts care to chip in.
So long as the charge is uniformly distributed on its surface (usually in practice because it is a conductor).pager48 said:Then it does not matter if the sphere is hollow or solid to be a capacitor?
It is my understanding that the Earth surface and the ionosphere form a concentric spherical capacitor that is largely charged by thunderstorm processes and this gives rise a ~uniform static field of several hundred volts/meter on average. The capacitance of that system should be roughly your number times 100 (i.e. REarth/ionosphere altitude). Also the outer surface is actually the ionosphere !lesaid said:It may be interesting to know that the Earth itself as an isolated sphere has a capacitance of about 710 uF.
lesaid said:You have the correct formula for the capacitance of an isolated sphere - i.e. ##C=4\pi\varepsilon_0 R##. Plug in the following values
##\varepsilon_0=8.85\times 10^{-12} F m^{-1}## (the 'permittivity of free space')
##R=0.1 m##
This comes out to ##1.11\times 10^{-11} Farads## or ##11.1\times 10^{-12} F## which is to say, ##11.1 pF##
lesaid said:Can you share with us what you are trying to achieve here? I have the impression you are wanting to design or build something?
lesaid said:But I would guess, not by a lot. It's something I'm now interested in figuring out, though I haven't got time at the moment!