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I've been reading the Feynman lectures, and in this lecture about lightning, he describes how the Earth is negatively charged, and how there is thus an electric field of about $100\ \tfrac{\text{V}}{\text{m}}$ near the surface of the earth. Here was the diagram to explain

To me, it seems like $100\ \tfrac{\text{V}}{\text{m}}$ could provide a lot of energy, a lot of small appliances run on $9\ \text{V}$ batteries, and the power point provides $240\ \text{V}$, which would be the potential difference between $2.4\ \text{m}$ of air, not all that much. There are probably problems in attempting to harness the electric field around the earth, but if I were to design a simple machine to harness the energy, how would I do so?

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QCD_IS_GOOD
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    Don't confuse electromotive potential with energy or power. The issue here is that we need to consider the resistivity of air, which let's say is $~1.3e16$ $\Omega / m$ which comes out to roughly $7.7e{-15}$ $A$ or $7.7\,femto-amps$ and $7.7e{-13}\,Watts$ of power available per vertical meter. (although we could also consider horizontal areas too) That's not much available power, even if you could somehow "pull it out of thin air". Your best bet might be to try to charge a very very wide and tall capacitor, but I'm not sure what the details would be. Certainly an entertaining question! – Falimond Oct 16 '15 at 06:08
  • @Falimond is the conductivity of air really the problem at hand? I could simply just have a structure made from copper that would acquire a great potential across its ends with no air present, and with coppers low resistance pull a lot of power if the resistance of air were really the main problem – QCD_IS_GOOD Oct 16 '15 at 06:11
  • I believe there's some effect of shielding due to the high resistivity of air that you need to consider. Briefly scanning the lecture, it sounds like this effect is largely, if not completely, due to ionic charging at the different strata in the atmosphere. When you're standing 2m tall, you're displacing the ions that would have taken up the volume where your body stands, and there's no potential difference from your head to your toes. – Falimond Oct 16 '15 at 06:29
  • Consider people and things being struck by lightning. This only happens when there's dielectric breakdown, and oh boy does current flow in that situation! Again, I think there's something critical here having to do with shielding. Hopefully someone else can chime in. – Falimond Oct 16 '15 at 06:31
  • It sounds like something Nicola Tesla tried to do in the early 20th Century but failed [https://en.wikipedia.org/wiki/Nikola_Tesla]. – Fred Oct 16 '15 at 12:16
  • I am highly surprised - I am nearly sure this $100 \frac{V}{m}$ is simply false (or it is bigger million times as the real value). – peterh Oct 16 '15 at 14:37
  • @peterh this article (http://hypertextbook.com/facts/1998/TreshaEdwards.shtml) gives various sources of measurements of the electric field, all of which give a value of roughly 100V/m – QCD_IS_GOOD Oct 16 '15 at 22:04
  • @JoshuaLin It would mean that a suspended dipole-charged steel rod would automatically set in a vertical direction as a compass... I've read this article but it seems to me very... unrealistic. Somewhere I've read there is a 150V of total potential difference between the surface of the Earth and the outer space. – peterh Oct 16 '15 at 22:11
  • @peterh There could very well be 150V potential difference between the surface of the earth and outer space and STILL be an electric field of 100 V/m close to the surface of the earth, because you have to take into consideration the effect of the ionosphere. According to (http://plasmasphere.nasa.gov/), the ionosphere is itself positive, and hence a stronger electric field would form between it and the earth. As to the dipole steel rod, by some back of the envelope calculations we can see that the torque on the induced dipole for a steel rod on the order of ~1 metre long is 10^-6, very small. – QCD_IS_GOOD Oct 16 '15 at 22:31

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I think Failmond's first post does a solid job with the analysis.

You have a voltage potential across a resistor(the air). You can not remove air from the equation as it is what is sourcing and sinking the voltage potential in question. The maximum energy you can harness is what it is already sinking.

The amount of power harnessable is indeed very low, but if I were to build a theoretical device to harness this power it would involve high altitude balloon system with lots of surface area. High altitude would mean high potential and higher the surface area the higher the current. It would be tethered with a light weight insulated conductor. The conductor could be very small gauge wire because of the very very small current, but the insulation would be very thick and have a high dielectric strength. Once the power made it to the ground it would connect to one leg of a specialized high voltage dcdc converter involving a spark gap and a transformer to convert the power down low enough to be useful. The other leg would be connected to an array of small copper wires going across the ground in all directions.

So, its possible, but at the end of the day you would produce a continuous 100 watts of power and have incurred millions/billions of dollars in expense. You have also created a really good lightning rod which would destroy the system unless even more expensive measures were taken. On a similar note, you can not harness power from lightning because it will destroy any reasonably sized system. And an unreasonably sized system, sized to handle 99% of lightning strikes without failure, would be many many many times over non-cost effective.

Its not reasonable to harness this knowledge for power production. However, in the future it could possibly be used for sensors much like an electroreceptive fish. https://en.wikipedia.org/wiki/Electroreception#Electrolocation

ericnutsch
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