For Woopy: Explanation for why you lose 1/2 energy in a capacitor

[TinselKoala]

http://www.youtube.com/watch?v=loSPjirgPhE

[MileHigh]

ibpointless:

You are mixing up what happens when you short two capacitors together and the electrolytic capacitor self-charging effect.  They are two separate things.  Have you looked up the effect on Google?

I am pretty sure that the self-charging effect is related to a natural potential difference created by the thin membrane layer(s) of the electrolytic oil and the availability of positive and negative ions in the air.  The electrolytic capacitor leads pick up charge from the atmosphere.  So you can say that the sun is the ultimate power source for the self-charging effect in electrolytic capacitors.

MileHigh

[TinselKoala]

ibpointless:

You are mixing up what happens when you short two capacitors together and the electrolytic capacitor self-charging effect.  They are two separate things.  Have you looked up the effect on Google?

I am pretty sure that the self-charging effect is related to a natural potential difference created by the thin membrane layer(s) of the electrolytic oil and the availability of positive and negative ions in the air.  The electrolytic capacitor leads pick up charge from the atmosphere.  So you can say that the sun is the ultimate power source for the self-charging effect in electrolytic capacitors.

MileHigh

Which can be significant. I left that 70000uF cap sitting disconnected overnight and today when I went to put it away I shorted the terminals with a keeper and there was enough self-charge that it gave a fair little spark when I made the contact.

A neat demo I sometimes do is to use a plastic pill bottle or film canister to make a capacitor by lining the inner side and the outer side with foil tape, then making a little spark gap at the top bridging the two "plates", and a frayed out fringe of braid or stranded wire to one plate to act as a pickup. This can be easily charged with a piece of PVC and some fur,  or by holding it up to a CRT monitor . The "self-charge" spark can get to 1/4 inch or more from just that simple little arrangement. Of course the charge doesn't stay on an improvised cap the way it will on a good electrolytic.

[NerzhDishual]

Please Gentlemen:

Do not use any resistor.
Just replace this resistor with a small motor geared to a small K7 recorder
mechanical counter (acting as a load).

Actually, in this case(?), you do not loose any Energy.
500  plus 250  plus  250 = 1000. (turns)

Very Best,
Jean

[Magluvin]

With resistive wire and non-ideal capacitors, each cap will have 5V.

With ideal wire and ideal capacitors (made from ideal conductors and ideal dielectric), each capacitor would have 7.07V.

However, does or will ideal conductors (zero resistance) and capacitors ever exist? Not likely.

But let's imagine that the day has arrived, and we have produced the ideal wire and capacitor (and inductor), and there is 0 inductance in the circuit. The time required to transfer the charge from the charged capacitor to the discharged capacitor would be infinitely small (0 seconds), and the current would be infinitely large. Can infinity really be reached?

With an "infinitely-large" current flowing for that "infinitely-small" amount of time, anything ferromagnetic or electrostatic within "infinite" distance from this "event" would be adversely affected by the infinitely-large electric and magnetic fields produced, including the iron in your blood.

In other words, you'd likely destroy not only yourself, but the universe as well. 

Hey Poynt

Remember we discussed some of this. So how does the inductor and diode, used to try and transfer all from cap A to cap B, seem to overcome most of the resistance losses? Is it because of the odd voltage divisions in the circuit due to the inductors working characteristics?

Like if we have cap A and B.  A 1000uf 100v  and B 1000uf 0v. We connect them but with an inductor between 2 legs of the caps , like all in series loop. Assuming the lowest resistance, room temp, what would be the highest level of voltage, peak, seen in cap B once the oscillation begins? 

Also, during the oscillation, same circuit, when the caps are at equal voltage in time, are they above 50v each? Lets say we are sampling the first cycles, from cap A to cap B, then back to cap A.

Mags