MH's ideal coil and voltage question

[Magluvin]

And they do, at 7.07V each, not 5V each.
Try stepping outside the box a little. The source cap can discharge to ANY voltage between 9.999V and 0V, depending on what it is connected to and how. The charge cap can charge to any voltage from 0V to 7.07V (max in this case). Can you not add a fig.4 to your drawing with an imbalance to create 7V on each cap?

I know you can (hint, use fig.3 and drop 4 more electrons from the top plate to the bottom plate). See, now we have 7V on each cap, and we didn't have to add any electrons.

If we have an ideal 10uf cap and a real 10uf cap,when we charge each to 10v, the electron number imbalance for each should be the same. So that is what we start with considering the electron imbalance.   From there on in, the Pos plate has no connection to the Neg plate for either cap.

So we only have a specific number for each cap imbalance. And when we do the cap to cap, that imbalance can only be divided by 2 when we let the two caps equalize. There is no chance in the 2 caps for there to be any more electrons taken from the top to the bottom, only from top to top and bottom to bottom(left to right, etc). All of the top electrons are trapped in the top side of the caps and the bottom side electrons are trapped in the bottom. Cap to cap that is.

For some odd reason(it seems) I see it clear as a bell. Im struggling to figure out better ways of explaining it.

Like as you said, take 4 more electrons from the top to the bottom, 2 electrons for each cap. Well after we are done charging the left cap, and then we do the cap to cap, when is it that those 4 electrons get a chance to move from the top to the bottom? Where is the bridge that they cross? The dielectric? At what point does this happen?

Mags

[Magluvin]

Sorry. Edited the first line and added the last question to my last post

Mags

[poynt99]

Like as you said, take 4 more electrons from the top to the bottom, 2 electrons for each cap. Well after we are done charging the left cap, and then we do the cap to cap, when is it that those 4 electrons get a chance to move from the top to the bottom? Where is the bridge that they cross? The dielectric? At what pint does this happen?

Mags

Right, 2 electrons per cap from top to bottom to change it from 5V each (15T, 25B) to 7V each (13T, 27B), agreed?

I don't understand part two of your question; We still start out with 10V on the source cap, and the electrons distributed as drawn (10T, 30B). However, now that we have ideal caps and an ideal wire connecting them, there is no loss, so after the transfer takes place, each cap will be left with an electron distribution of 13T, 27B.

[Magluvin]

Gotta git.   Im not getting what you are getting at.     If we look at the 2 top plates after cap to cap, the electron imbalance can only be half in each top plate of the left caps total electron imbalance after it was charged. As would also be with the bottom plates. There can be no more imbalance changes between top and bottom after the initial charge of the left cap.

If we were to move more electrons from one top plate to the 'other' top plate so that the 'other' top plate has 7 electrons, for just say 7v,(do same for bottom plates) then they would not be equalized in electron count and the other top/bottom plate would only have 3 electrons. When we talk about equalization of the 2 caps, Im talking about equally divided by 2 distribution. I dont see any other way of the electron distribution other than divided by 2.

If you have a pic editor, could you show me how and when the electron changeover is done how you say?  Im having a hard time getting it. I used Picpick. Free and fairly simple to use. Mostly use for screen grabs.

Mags

[tinman]

For Pete's sake Brad. This is simply a case of different terminologies used, by you and by MH, OK? Some folks do understand what it means, put the way MH put it, and some don't. Can we settle this now and never argue about it again? MH did not mean that the voltage would change after it was set, he meant that it can change at various times between t=0 and t=x. In other words, he meant that it would never deviate from its setting until the next setting. But the fact that it is being changed a number of times between t=0 and t=x, means it IS varying over that time period. OK? THAT is what he meant. Don't think that MH doesn't know what an ideal voltage source is, that would be ridiculous of you, and this silly argument is becoming extremely old.

I'm not going to comment on the last part of your post as it is simply not worthy of comment. Same with MH's knee-jerk reaction to Mags' last few postings. I think you both need to take a pill. 

No,MH wrongly accused myself and others that we do not know the definition of an ideal voltage source. He clearly stated that we say that the voltage cannot change with time-and that is crap. 300 posts ago,i posted my understanding of an ideal voltage ,and it is correct,and i will not have MH bullshitting about me being wrong-full stop.
The voltage can only change if determind to do so by the user-fact.
So for MH to say i am wrong,is just bullshit-plane and simple.

Why have you not told MH that he is incorrect,in stating that i do not know what an ideal voltage is?