Another tough 'elementary' question from my tutor

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Hi Dear jasmin_123!

My Previous Explaination was a mathematical one. But if you want a physical one I can say that an infinite current can't pass through such physical circuit, so the wire may burns out of there may be some electric discharge you may see as flares, when you connect the wires. I've done it. You can test it too!

Thanks!
 


Why would ALL the electrons pass to the second cap initially? and why would this continue as a sustained oscillation over time and into steady state? What causes the oscillation?
 


I don't know how you come up to the conclusion that the charge will oscillate between the two capacitors.

What will happen is at steady state the voltage of the two capacitors should be equal because they are connected in parallel. For the voltage to be equal the charge stored in both capacitors must be equal since V=Q/C. Therfore the initial charge in the first capacitor will be divided between the two capacitors. And the charge stored in each capacitor is Q/2.

But this charge movement involves enery loss. As explained by jayc in this link
url]h**p://&highlight=
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Some work must be done to move electron charges to plates of different electric potential.

If the circuit involves a resistance then there will additional energy loss due to the resistor. But in your case the resitance is 0.
 

Oscillation will occur only if there is inductance between the capacitors. I posted this on Feb. 10:
 

Learner said:
Why would ALL the electrons pass to the second cap initially? and why would this continue as a sustained oscillation over time and into steady state? What causes the oscillation?
Click to expand...

1) The five electrons in the first cap have a potential energy.
2) Zero resistance cannot pass the charge in portions, thus, al the electrons will pass to the second cap.
3) Their potential energy will be transformed, first, into kinetic energy and then into potential energy when they reach the second cap.
4) …

The oscillations are caused by the energy transform from one type into another.
 

What if.. there was a serie resistor between the caps?
How the energy would be dissipated? All in R? charge displacement W?

and what is the limit when R-->0.
 

Hi Jasmin,
If the wires connecting the the two capacitrs have zero resistance each wire can be shrunk to a point and in this case is it not necessary that the potential across the the two points should be same at any given time. If at any time each capacitor should hold different charges will it not violate this basic requirement?
Regards,
Laktronics.
 

Mr.MEB said:
and what is the limit when R-->0.
Click to expand...
For R-->0, the limit is: a half of the energy is dissipated.
But for an R=0, no energy is dissipated. This is singularity!
 

Hi jasmin_123,

Why do you state that “there are NO energy losses”?
Conservation of charge is not necessarily compatible with conservation of energy!
MrMEB’s proposal is the key:
Imagine that the two Cs are connected with a resistor. At steady state (ok, it woult take an infinite time, but it does not matter) the charge is the same on the two Cs are at V/2 and half of the energy is lost in the R. This happens regardless of the value of the resistor.
Now imagine that the same experiment is repeated with a smaller R. The peak intensity increases, the time constant decreases, but the lost energy is the same (V^2*C/4) regardless of R and the “final state” is the same. The value of R is relevant only in how fast the system approaches that state.
Now imagine what is the limit as R goes to 0. The dissipation is the same (infinite intensity, zero time), but the limit exist!

Regards

Z
 

laktronics said:
Hi Jasmin,
If the wires connecting the the two capacitrs have zero resistance each wire can be shrunk to a point ...
Click to expand...
But what about superconducting wires
---
Can you, gentlemen, think of something abstract???

Added after 5 minutes:

zorro said:
Why do you state that “there are NO energy losses”?
Click to expand...
Hi, zorro,

For R-->0, the limit is: a half of the energy is dissipated.
But for an R=0, no energy is dissipated. This is singularity!
---
It was a 100% abstract question. Of course, such a system does not exist in nature.
 

jasmin_123 said:
2) Zero resistance cannot pass the charge in portions, thus, al the electrons will pass to the second cap.
Click to expand...

What is this piece of info based on? Can you show some reference? eg. science/physics theory/facts online etc...

Or explain why would a bunch of electrons flow as a single entity instead of individual particles?

Example
There are 2 identical tanks, no.1 filled with water and no.2 is empty. They are then connected together by a pipe at the bottom of the tank so water could flow, would the water in tank 1 be drained to fill tank 2 and then tank 2 be drained to fill tank 1? and this process repeates itself? (Assume that pipe has no friction)

Why does this example not reflect the capacitor scenario?
 

Learner said:
Why does this example not reflect the capacitor scenario?
Click to expand...

It does reflect if the pipe has zero 'resistance' and passes all the water from one tank to the other.

Why a zero-value resistor conducts all the charge simultaneously? Because its conductivity is infinite!
 

Hi Dear jasmin_123!

I have followed the subject and I got that there is no clear target to getting conclusion. I mean you'd better say if you want the problem to be solved according to Physics, Mathematics, Electronics, etc. Please say the way you like the problem to be explained.

Thanks!
 


Energy in 1st capacitor:

W1 = 1/2 * C * V^2

Energy in 2nd capacitor:

W2 = 1/2 * C * 0^2 = 0

Energy in this closed system: W = W1 + W2 = W1


Energy DOES NOT CHANGE, therefore:

Energy W = 1/2 * C * V^2 = 1/2 * C2 * V2^2

where C2 = 2*C (C connected in parallel) and V2 is the requested final voltage, then

V2 = V * sqrt(C/C2) = V / sqrt(2), so simple...

Best regards,
Eric

Added after 3 minutes:

abel51 said:
... bla bla ...

Half of the energy Gone!!
Click to expand...

are you serious ?

Eric
 

Hi, Eric,

Have you heard anything about the charge conservation low?

Cheers!
Jasmin
 

Ideally, the answer should all be the same regardless if its in math or theory. They are just in different "language".

Assuming in the case of super conductor,

Superconductor: An element, inter-metallic alloy, or compound that will conduct electricity without resistance below a certain temperature. However, this applies only to direct current (DC) electricity and to finite amounts of current.
Click to expand...

**broken link removed**

So because of the nature of SC, the charges can not oscillate as you've proposed in the cap.

====================================================

In regards to the water tank analogy, with a modification.

Assuming there is no pipe but only 1 glass tank with a seperation in the middle, left hand side of tank filled with water and right hand side empty.
When you remove the seperation, water flow from left to right; there are oscillations in transient state but not in steady state. There is NO RESISTANCE to the water flow from side to side of the tank. Why doesn't ALL the water passes from one side of the tank to the other and leave 1 side empty? Why doesn't the water continue to oscillate in the steady state as you assumed to be when there is NO RESISTANCE to the water flow? (If you like, this can be inside a vaccum chamber)

To represent the cap example fully, we can place a mirror at the bottom of the galss tank. The mirror reflect whats happening in the tank which represent what happens on the other side of the cap.
 


Eric.

According to your calcultation there are more charges in the capacitors than there was in the beginning. This violates charge conservation law.

The stored energy in the beginning cannot be the same as the stored final energy because work is done to move electric charges between points of different electric potential. Work meanse energy loss. Read Coulomb's law.

W=qV, which is derived from coulomb's law. F=kq1q2/r²

F=kq1q2/r²
F=qE
W=F*distance
W=qE*distance
W=q*V

BR
 

It is not fair, folks,

It had been stated clearly, from the very beginning, that a lossless system was considered.
However, instead of answering the question for zero losses, you try and prove me that losses cannot be zero. I write you again: "consider a zero-losses case," and you respond: "look, Jasmine, losses cannot be zero."

Are you kidding?
 

Jasmin_123,

Please try to answer to this question:
How many energy dissipates an infine current passing through a zero resistance conductor?

You answer (up to now) is: “zero, because the conductor is lossless.”

The correct answer is: “the question is not well posed. The answer can be found only as a limit. The result can be zero, finite or infinite.”

Please see my post above.
Regards

Z
 

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