Hello, I made a Leyden jar out of a quart jar filled with a salt water solution, and connected it to a switch for safety.
But I need help on how to charge it. I tried 120VAC positive to the top, and negative to the foil outside. All connections are ghetto but they work and I got 120 VAC on the multimeter.
But when I (accidentally) touched the 2 ends of the jar, all I got was a tingle from my finger (top) to my thumb (aluminum foil). I'm not sure it worked and I searched and searched on Google but I get nothing. Can I use a wall outlet to charge my Leyden jar?
>Tl;DR
>Leyden jar
>120 VAC to charge or Nah
>inb4 'don't kill yourself'
>>969394
Leyden Jar is DC you're trying to charge with 120V AC power. It doesn't work that way. You can only store DC but the outlets in your house etc. are AC.
>>969394
you need to rectify the AC to DC if you want it to accumulate charge, that said, have fun electrocuting yourself
>>969406
How do I get 100-150 volts DC to the jar. Batteries? AC to DC conversion? (is this possible?) I do have a shit ton of 9v batteries though
>>969407
You either need a bridge rectifier, or 4 diodes wired into one, use the googles anon, they're pretty simple to make.
And again, given your lack of knowledge about electricity, have fun killing yourself.
>>969407
>100-150 volts DC
Don't even bother. You want 1000s of volts for these.
>Wimshurst machine.
>>969408
>killing yourself
Not enough current anon
Well I posted here to learn, and I am learning.
A Leyden jar is normally charged using a static electricity source such as a Wimshurst generator or Van de Graaf generator. These generate high voltages but minute currents (which isn't an issue as a Leyden jar can't hold much charge anyhow).
If you're only using mains voltage, you may as well just use a normal capacitor (you still need to rectify the mains).
A Leyden jar is just a very primitive capacitor with a large distance between the conductors. As a consequence of the distance, they have very low capacitance but are capable of withstanding high voltages.
Normal capacitors have a much smaller distance (like, microns) between conductors, a relatively large surface area for their size (having many layers rolled into a cylinder or folded into an approximate cuboid), and a dielectric (insulator) which polarises (this has the same effect as reducing the distance between conductors).
Essentially, voltage versus capacitance is a trade-off. Higher voltage requires a thicker insulator which results in lower capacitance.
