I get between .7 and 1.3 volts out of a 20 foot tree depending on the weather. I understand this is "telluric" electricity with pretty low amperage. Where is this charge coming from? The sun, right? Lets forget solar panels and tap into that!
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Just because there is a sizable potential difference doesn't mean there's a great deal of energy stored. As soon as you begin to tap that energy (if you could) you might find that the voltage drops off swiftly. If it's coming from the Sun, you'll probably find that not much of its energy is being stored in this way compared to what plants or our solar cells can "fix". – Selene Routley Feb 03 '15 at 02:59
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You're not measuring a charge with a voltmeter, you're measuring the potential difference due to a current of charge. The Earth can be neutral and still have ground currents. – Feb 03 '15 at 03:02
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Yes there is a net charge on the earth; no you can't use that to generate energy. For a current to flow, charge has to be moved by an electric field. In the case of Earth, how do you propose to create a circuit to tap into this field? You could move a charge from the ground to the clouds - and maybe do a very small amount of work while moving it. But then what? If you let go of your charge, and it flies off into the universe, the earth will be less charged. If you try to move it back to earth, you need to do work. The amount of net charge on earth is pretty small - compared to the energy of the solar radiation.
A bit of math:
The capacitance of an isolated sphere is
$$C = 4\pi \epsilon_0 R$$
For Earth, that gives a capacitance of
$$C{earth} = 700 mF$$
The net charge on earth (without counting the atmosphere) is about $10^5C$; the energy stored in a capacitor is $E=\frac12 C V^2 = \frac{Q^2}{2C} \approx 7 GJ$.
7 Gigajoules is not all that much. The US uses about 70 TWh / day, or about 3 TW at any one moment; it would take just over $2 ms$ to use up all that stored energy, and it would take a lot longer to replace...
By comparison, the total solar energy that hits the earth (at about $1 kW / m^2$) is roughly
$$E = \pi R^2 \Phi = \pi * (6.3\cdot 10^6)^2 * 1000 \approx 10^{20} W$$
That is energy that arrives every second of every day. Of course a lot hits the ocean (70%), and a lot hits places where there are clouds, or where solar panels cannot be built for one reason or another. But you are still many orders of magnitude above the 3 TW - or, if you want to power the entire world, 15 TW. In fact, according to http://en.wikipedia.org/wiki/World_energy_consumption, you could generate 50 TW of solar power - three times what the earth needs today, and still a tiny fraction of the solar power hitting the earth.
I think solar panels are still a better bet than the charge on the earth.
See also What is the net charge of the Earth?
