Energy Loss

[rpapps]

Sy is on the right track.
If you connect two equal capacitors in parallel the charge will be shared equally (Law of conservation of energy).
Now V=Q/C so, in a perfect world, if you double C, V halves.
No violation.
In the real world, some energy is lost due to thermal dissipation and electromagnetic radiation in the connecting wires so the end result is slightly less than V/2.
Physics 101 people!
Reference:
Physics for Scientists and Engineers
Fourth Edition
Raymond A. Serway

[ChocoHolic]

Originally posted by rpapps
Sy is on the right track.
If you connect two equal capacitors in parallel the charge will be shared equally (Law of conservation of energy).
Now V=Q/C so, in a perfect world, if you double C, V halves.
No violation.
In the real world, some energy is lost due to thermal dissipation and electromagnetic radiation in the connecting wires so the end result is slightly less than V/2.
Physics 101 people!
Reference:
Physics for Scientists and Engineers
Fourth Edition
Raymond A. Serway

rpapps:
Our trouble in the moment is that double C and half voltage DOES violate the law of conservation of energy. Half of the original energy is lost.
Which is also plausible, because if split two double C and half V the unchanged amount of charge is placed on lower potential. Less energy.
The funny thing that we are loosing the energy even with a theoretically lossfree charge changing model.

If you want to split the energy onto two caps without energy loss, you would need an addiotinal ideal choke and some ideal switches.
...resulting in a nice energy transfer and then the voltage in each cap would be something like 70% of the original volatge.
This resonant event would nicely match to the law of conservation of energy, but would somehow violate the law keeping the amount of charge constant.... charge is then more afterwards, but we did not supply external charge... strange world !

[rpapps]

Sorry chocoholic, I don't see it and I can't read your scans to verify the math.
Cheers
Rob

[ChocoHolic]

Our discussion is based on the following phenomenon:

Imagine two caps, both 1F.

In the beginning Cap1 is charged to U1=2V.
Cap2 is uncharged:
Total energy in the beginning E1+E2 = 1/2 x 1F x (2V)^2 + 0 = 2J

After connecting the second in parallel, both caps are charge to half of the voltage:
Total energy in the end: E1+E2 = 1/2 x 1F x (1V)^2 + 1/2 x 1F x (1V)^2 = 1 J

Half of the original energy is lost.

[ChocoHolic]

Originally posted by SY
Here's an analogy that (at least for me) clarifies things. Imagine your first situation, a cap with a voltage V and a charge Q. Now, move one of the plates closer to the other to increase the capacitance...

Hi SY,
I think in your example things are less strange.
Due to opposite charge on both plates there is mechanic force, which is pulling together both plates.
If we now make one plate movable (and free of friction), we could use this force to pull something up. Mechanical energy can be calculated by increased mechanical potential energy or by integral of F (ds).

[rpapps]

Hmmm
I shall have to think about this.
Hopefully not for 7.5 million years.
Bedtime here, back tomorrow.
Cheers
Rob

[Joules]

Dude --- You keep assuming the voltage will be cut in half by adding a second capacitor in parallel, but there is no law to prove that it will.
The law of consevation of energy is real. so based on that, the problem is

> 1 capacitor - .5(1F)*(2v)^2 = 2 joules

> 2 capacitors - 2 joules/(.5(2F) = sqrt of 2 or 1.414 Volts shared on bothe capacitors.

a simple reverse check > .5(2F)*(1.414v)^2 = 2 joules

[SY]

Y'all have to work harder on this one...

edit: ok, I'll give you yet another way to think about the problem. Your capacitor is charged, then we insert a chunk of dielectric with k = 2.

[ChocoHolic]

Originally posted by SY
Y'all have to work harder on this one...

edit: ok, I'll give you yet another way to think about the problem. Your capacitor is charged, then we insert a chunk of dielectric with k = 2.

How about answers, not questions?
I would guess that in this case the voltage will drop to 70.7%, but that's just a guess.
Let us know your view.

P.S.
No comment on my 'integral F ds' from your previous question?
At least your own answer to this question might help us.




Joules:
Besides conservation of energy, there is also a law of constant charge... Both laws seem to contradict here.

[poobah]

Alright SY... poop rolls downhill and only downhill.

rpapps... the rub here is that V = Q / C, & E = Q^2 / (2 * C)

In this example charge must be conserved, energy as well, no one said the enregy had to stay in the caps though.