I don't know why this is being discussed again because I already discussed it with Magluvin on the other thread.  If you can imagine a cap transfer where no energy is lost and the two caps are at 7.071 volts then you need more charge than is in the original cap sitting at 10 volts.  An idealized arrangement with a coil and a switch can do this.  Without the idealized arrangement the initial amount of charge in the original cap loses some energy through resistance when it spreads itself out between the two caps.

Conservation of charge equals conservation of momentum in the two colliding masses example.

 Energy stored on a capacitor has counterintuitive elements.

Why would that be John?

I think he wrote that, because two electrons can represent different energies depending on their separation distance and the work needed to move them closer together is a result of their repulsion force integrated over the distance of their approach. 
Reminder: The instantaneous repulsion force of two electrons is inversely proportional to the square of the distance (1/d²) between them.

BTW: this is an example how the same amount of charge (here, the two electrons) can represent very different levels of energy.
In other words: Charge alone, is not energy.

   Capacitor = potential .
 
   Sounds good to me.

Inductor = kinetic.

With the exception of MHs question,between T=3s and T=5s,where the magnetic field is stable,and so the stored energy is potential energy.


Brad

That's not right but more importantly how did you arrive at that answer?  That's the whole point for the brainstorming, to bounce ideas back and forth.

The other question is why are you talking about phase here and how do you define it?

I agree.

I agree.

Ok.

When I was saying to simplify things by saying that the electron count was -1000 electrons for the Pos plate and +1000 electrons for the Neg plate when the cap is charged to 10v, the 1000 number is just to simplify things.  The real numbers are probably so large, but the symmetry should be fairly close. Nearly as many electrons come out of the Pos cap lead as there is going in to the Neg cap lead.

Would you agree that if we charged the cap from 0v to 10v, and we 'could' count the numbers of the electron count offset between the Pos and Neg plates, then if we discharged the cap to 0v and recharged the cap to 10v, and we did another count of the electron offset between the two plates, would that number be the same as the offset count after the first charge?    Im talking theoretically 0v and 10.00000000000000000000000000000000000v Not a 10.001v or 9.999999v.  0v to 10v.


Mags

Agreed. I think you may want to skip ahead; see my last response to your questions.

"Well we can't have ideal conductors (when shorting an ideal cap or voltage source), so this scenario is unachievable/unsolvable to begin with. You would have infinite current as soon as the two caps are connected."


"Assuming the universe didn't blow up trying to source infinite current, the voltage would be 7.07V in each cap. The electron balance would be 707 on each plate, vs 1000 when it was 10V, assuming there is a valid linear relationship between the charge count and voltage of course."

Ok. Good.   

1 cap at 10v -  +1000 excess electrons on the Neg plate and -1000 electrons on the Pos plate.

1 cap at 0v -   each plate has an equal amount of electrons and no imbalance.

 If we let the cap discharge into a 0v cap and equalize, how do we get +707 offset count on both caps Neg plates and -707 on both caps Pos plates?  If we divide 1000 by 2 we get 500 + and - for each cap.  Where did we get the extra electrons to get the 207 count difference?   Its like saying if we have 1 gal of water and we divide it equally between 2 containers that each container would have .707 gal of water at no loss, but if we ended up with .5 gal per container that we have lost half. In a way.
 The total offset for the 2 ideal caps would be 1414 electron offset count.  We started with 1000 offset in the source cap. Can you see the discrepancy?

So how do we account for the extra electrons in the ideal cap to cap?  I can see how we do it with the inductor, where we cut off the source cap at 7.07v, then let the inductor freewheel electrons from the Pos of the receiving cap to the Neg until it reaches 7.07v.  All say done with super fast timed switching, no diodes..

But with the cap to cap, there isnt the option of gaining extra electrons in the offset to alter the original count. They can only be distributed equally between he 2 caps. 1000 + and - offset count of the source cap to 1000+ and - offset count for both caps total, 500 + and - offset each.   The above explanation with the inductor, we stopped the flow from the source cap at 707 + and - offset count, and at that time the inductor pumps the receiving cap the rest of the way by way of taking from the Pos and giving to the neg.

Similar to the air tank with an air motor with fly wheel.  100lb air on source tank and 0lb in the receiving tank. Open the valve and the pressure gets the flywheel going while the receiving tank fills.  We cut of the source tank at 70.7lb and switch over to letting the flywheel pump 'extra' air from the outside, till the receiving tank is at 70.7lb.  If we simply did tank to tank, we would have 50 lb per. Heat losses or not, tank to tank could never be a result of 70.7lb per tank all said and done.


So if we can agree on the simple mathematics of the offset count, then we would have to agree that the ideal cap to cap would result in 5v per cap. So where did we lose the 50%energy with no resistance and or heat losses?   Thats my point Poynt.

Mags