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Tesla patented a device for gathering energy from light, using the photoelectric effect. (US 685,957 - Apparatus for the Utilization of Radiant Energy):

Telsa's photoelectric effect generator

Basically just a sheet of "highly polished or amalgamated" metal (top) connected to a capacitor (the ⊔⊓ shapes) connected to the Earth. (The other stuff on the right of the image is a load intermittently driven when the charge on the capacitor rises high enough.)

Would this actually work? Why or why not? Does it need to be in a vacuum? Would the plate stay charged after electrons are thrown off, or would they be attracted and drift back to it and neutralize it? How does the capacitor to earth make this better than a plate hanging in space?

The patent demonstrates that he doesn't understand the cause of the photoelectric effect ("sources of such radiant energy throw off with great velocity minute particles of matter which are strongly electrified, and therefore capable of charging an electrical conductor"), but he understands the results.

(Incidentally, I have tried building this with a piece of aluminum foil, electrolytic cap, and wire strong enough to stick a few inches into the ground. I didn't see any voltage with a multimeter in bright sunlight.)

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Extra image is not necessary, but you may want to add the description of the middle device. This image is very unclear. – hwlau Dec 8 '10 at 9:26
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Exegese of that "middle" device is rather easy, it is an electromagnet coupled to a cam wheel by a ratchet drive. This will make revolutions going on in the same direction although the armature of the solenoid makes a reciprocating movement. The thing more to the right side is a electrostatic switch I suppose. The idea is that those two leaflets hanging down like in a electrometer will contact after some potential is reached. – Georg Jan 31 '11 at 16:40
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Electrolytic capacitors have low parasitic resistance (leakage current) and a high capacitance. These two features combine to prevent you from seeing any voltage (it will leak off too quickly). Follow the advice of Tesla and go with a high quality, high voltage capacitor. I suggest teflon. Make sure that your voltage measuring device doesn't short out the capacitor. To test this, charge the capacitor with a voltage source (be careful to not kill yourself, use maybe 12V), and see if you can detect the charge. – Carl Brannen Jul 13 '11 at 11:18
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What Tesla was seeing was indeed the photoelectric effect. This is why he thinks that the beam consists of "positively charged particles". So it helps to use a metal with a low work function. This is a problem as the good metals are too reactive. But the shortest wavelength hitting earth is 100nm or around 12.4 eV and this is enough to knock electrons off of any metal. – Carl Brannen Jul 13 '11 at 11:27
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Finally, it seems to me that you're restricted to the voltage of the electrons, i.e. about 12.4 - 4 = max photon - aluminum = 8 volts. I don't see how he got extremely high voltages. Maybe I misread it and he only got high charges (by waiting a long time). – Carl Brannen Jul 13 '11 at 11:33
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The energy needed to remove an electron from a solid is called the work function. For most metals you would need UV photons (300 nm for Aluminium) that rarely reach the Earth's surface. Visible light can eject electrons from alkali metals, but the quantum yield (the probability of electron emission per incident photon) for pure metals is low (probably less than 1%). Materials like CsTe that are used in photocathodes have efficiency up to 40% (at certain wavelength) but they are expensive and difficult to handle in open air. Silicon solar cells also utilize photoelectric effect and compared to metals they are efficient and inexpensive.

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I would say that it is an issue of efficiency. In principle it could work, but there are more efficient ways. – Vagelford Dec 8 '10 at 9:33
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Not exactly. For visible light the efficiency of most metals will be exactly zero because the photon energy is just not high enough. – gigacyan Dec 8 '10 at 9:46
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So it is a issue of efficiency. – Vagelford Dec 8 '10 at 10:47
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An efficiency of exactly zero is not really "an issue of efficiency". – endolith Dec 8 '10 at 21:05
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If it was exactly zero, you would be right. But as you say in your post it isn't. – Vagelford Dec 9 '10 at 9:56
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