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Can i charge a capacitor fully with only static electricity? I already took a look at: Electrostatically charging a capacitor and How to charge a capacitor with static electricity? but i'm not sure this answers my question fully. Can you explain to me if it is possible to charge a capacitor from for example a wimshurst machine? What needs to be taken care when the capacity of the cap changes (i.e. if one time i have 100nF and a different one has 1000uF)? Im trying to charge a homemade foil capacitor and before building a bigger one I would like to know if doing this is feasible and what specifications i should aim for. By that approach I want to avoid building a voltage multiplier or other high voltage supplies.

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    \$\begingroup\$ Very interesting question because it brought me back, a bit. Google easily found this link: Electrostatic Machines. That link is a compendium of various machines, many of which the author has built and experimented with. Worth going through (at least, to me.) \$\endgroup\$
    – jonk
    Sep 3, 2018 at 20:20
  • \$\begingroup\$ On the wimshurst site there is this program: coe.ufrj.br/~acmq/programs/wmd.zip . I tried it out but im not yet sure what to do with those numbers. Could you maybe explain to me how I could go from these outputs to a calculation for charging a capcitor? For the sample in the program it states it outputs 17.08uA. Does this even help me with calculating the time? In another post I get I=VbRe−t/RC (for calculating how much amps will drawn with a fixed voltage supply). Im almost sure I cannot take this formula and just solve for the voltage that will be in the cap. \$\endgroup\$
    – ShinuSha
    Sep 3, 2018 at 20:56
  • \$\begingroup\$ That program doesn't accept specs for caps (or Leyden jars.) However, having them with larger values provides heavier-looking sparks. Much depends on the environment (humidity, atmospheric pressure, surrounding gas type, etc) that I don't think you will get an accurate answer about timing without a lot more experimental control/detail and a heavy amount of mathematics and the application of several physical theories to the problem. See, e.g., physics of gas discharge. \$\endgroup\$
    – jonk
    Sep 3, 2018 at 21:18

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Sure. In your wimshurst machine example, you're already charging up two capacitors that are built into the machine, they're the jars on the right and left. enter image description here

If you want to charge up another capacitor, you could connect it between the rod that's going into the jar and the outer bit of metal (though you'd probably need a capacitor for each jar), or charge it by putting it between the balls. However, you need to make sure the capacitor is rated to the voltage you're charging it to (on the order of 10kV), which is not common. Since you're making it yourself, you will need to use a thickish dielectric and be prepared for it to arc over. You also need to be prepared for it to take a long time to charge. According to wikipedia, a leyden jar has about 1nf of capacitance, so if you add say a 10nF capacitor to each jar, the sparks will occur about 10x less often.

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  • \$\begingroup\$ Thanks helped alot. I know the voltage the machine can produce at max is restricted by multiple factors (i guess the plate size, plate insulation, etc.). Is there a site or some material that describes how to 'calculate' the voltage expected? + How would I calculate how long it takes to charge a capacitor with x capacity? Is charge 'created' linearly by turning the machine? \$\endgroup\$
    – ShinuSha
    Sep 3, 2018 at 19:56
  • \$\begingroup\$ The time taken would depend a lot on your exact setup. For a rough estimate you could take the time taken to charge the leyden jars and multiply it by the increase in capacitance. The voltage is going to be limited by your spark gap and the breakdown voltage of the air, but it can be less if you have a lot of corona discharge (sharp corners). This answer gives a rough figure of 1kV/mm of gap. \$\endgroup\$
    – C_Elegans
    Sep 3, 2018 at 20:04

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