Conservation of Energy applied to a potential difference.

[silverdiesel]

I am looking for some help on how to apply the law of energy conservation when applied to a voltage difference. Specifically, here is what I am thinking.

If you build a voltage difference on a capacitor, I know it takes energy to build that difference, but once it is there, can the resulting electric field preform work without reducing the potential difference? Say I have a charged sphere on a lever connected to a weight, or gears, some load. Then, I place this sphere/lever system into the electric field of the capacitor. The sphere will move, as will the lever, as will the load. Now, if the lever is not long enough, or not allowed to move far enough to actually touch the plates of the capicitor, there will be no loss of energy in the potential difference, yet, the load will have moved by some use of energy from somewhere... right? What am I missing? I hope I have explained my set-up well enough to be understood.

 

[erickalle]

there will be no loss of energy in the potential difference,

Yes there will.

By inserting a metal object between capacitor plates you effectively make the capacitance bigger. You can see that by letting your object fill almost all of the space between the plates without actually touching the plates. You have now 2 capacitors in series, which combined are bigger then the first one.
Since pd = q / c and c is now bigger so pd is now smaller. If you had an extremely sensitive volt meter you could measure this difference. Perhaps an electroscope could show the difference.
Even if you would insert a plastic object you could make a roughly similar argument because then er changes so that the capacitance also gets bigger.

 

[silverdiesel]

okay, that makes a lot of sence. Thank you for clearing that up.