It appears the Earth has a positive electrical charge and the atmosphere is negative.
Can this be demonstrated by wiring a long vertical wire to the negative terminal of a 9V battery to see if the superimposed potential will tend to neutralize its charge, as measured on a voltmeter?
Is there any way the timeframe for full depletion of the batery could be calculated/estimated?
Bob Lipton
snipped-for-privacy@radiodyne.com wrote in news:e19os7psc4s00jinrr5d57e22vvsqulqq6@
4ax.com:
You can use the voltage difference over about a 100 m altitude, to run an electro-static motor( google )
Hmm... decades.
The atmosphere isn't very conductive, so the current flow into even a very large metallic structure is very small (microamperes). The average 9V battery has 200mAh or so capacity, so will be depleted by 1uA in around 22.8 years.
Yes the atmosphere is charged, but that doesn't mean you can draw power from it. It's like saying, yes, a spring holds a force, but you can't just look at it to get power out of it. You have to cut in and complete the circuit to discharge it.
That said, if a space elevator is ever built (of whichever type: centripedal belt, "beanstalk", or what have you), and it penetrates the ionosphere, it has the capability of interacting with this charge, because the ionosphere is appreciably conductive. Billions of volts would be available along its length, at a pretty fair current I'd imagine (amps?), at least until the ionosphere is discharged to equilibrium. This would probably alter global weather patterns! I don't know if anyone has considered or analyzed this effect in a space elevator proposal.
Tim
But I believe the "Earthing" of the antenna at the battery closes the circuit. And more than 1uA has been achieved. Look at Jefimenko's ES motors.
Bob
Hi Bob, look up an electric field mill.
Another instrument that I'd like to make some day.
George H.
Build your gizmo, and measure the current. Chances are that rain, fog, humidity, etc., will increase the current over that of a dry day, so take that into account in your measurement program.
I suspect the discharge will be sloooooooooow.
So THAT'S why batteries discharge over time when allowed to sit in the atmosphere !!!
Yes, but very poorly. You have to tap into the source -- the ionosphere -- to draw any appreciable current.
Skimming and searching for "micro" turns up three occurances: "such antennas produce about a microampere of current" (about halfway down). The figure given of ~1 pA/m^2 suggests 1uA is from a rather large antenna of 1000 x
1000 m.The next occurance says a sharp-pointed 60' wire could do ~1uA at 2kV.
The last occurance is talking about a microwatt motor, which is... powerfully unimpressive.
So, surprisingly, the number I heard tossed around a long time ago still seems to be reasonable!
Tim
The link is to a hodge-podge of information, involving 2 issues, although the distinction is not clearly made. The first is electrostatic motors, where the supplied static electricity may or may not be the earth's field. The 2nd issue is the earth's field.
It seems that Jefimenko did build some e-s motors with modest power (.1 hp). But they were run by generated charges.
As far as earth field power:
Dr. Jefimenko discovered that when sharp-pointed antennas are designed for a sufficient length to obtain at least 6000 volts of threshold energy, the fair-weather current density available is about a picoampere per square meter. Such antennas produce about a _microampere_ of current. [my emphasis]
It's not made clear how the square meter is measured, or why current is proportional to area (volume seems right-er). But that's typical of the site: numbers and ideas thrown around without real explanation. Superficial "science".
Bob
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Feynman vol.2 sec.9 lists the E field as 100V/m and current as ~10pA/mm^2
During a T-storm those numbers can change. :^)
George H.
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The voltage increases as ~100V/m so you've got a long wire sticking into the air. the pA/m^2 is the leakage current perpendicular to the field. (A square meter of metal a bit above the gound.) 1 uA from a 60' (20m) wire stuck up above the ground, is some complicated integral... I bet the voltage is a strong function of the current.
George H.
^mm is a small area, so, if one has a lot of plates, large ones, those could be connected together, something like a dense heat sink. That is, if I am reading that correctly. 10 Pico Amps per Square millimeter..
It seems back in my school days, I seem to recall something about
50 Volts or 50 feet for 100 volts.. its been a while ;)Jamie
Some sources claim a 300 kV potential difference between the ground and ionosphere and the global leakage current of the order of 1000 A. If this current is distributed evenly, the clear air current would indeed be a few pA/m², not very practical.
However, at certain conditions, with high wind and rain (snow/hail), conductors can be accumulate a lot of charge. Take for instance, a simple wire dipole antenna a few ten meters long, with one side connected to the center of the coaxial cable and the other to the shield with an UHF connector at the other end. When the UHF connector is not connected anywhere, there can be several flashovers a minute between the UHF connector pin and barrel. This electrostatic charge might even do some useful work.
Me, too.
The Langmuir Lab guys at NM Tech use those. Made in-house, nice machining job.
Easy to calibrate.
That may be an ideal demonstration of FREE ENERGY..
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