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I think it will be more when measuring it from the inside...many times more, and will be directly proportional to the distance between the plates.

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I think it will be more when measuring it from the inside...many times more, and will be directly proportional to the distance between the plates.

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Born2bwire

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A manifestation of this fact is that, if the arrangement is dismantled, the P.D between the dismantled plates will be much higher than P.D of the charge source.

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Born2bwire

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There is no electric fied inside of a conductor at steady-state

I'm talking about a parallel plate capacitor and the P.D from its inside, i.e not the way we usually connect the capacitor to the external circuit but the path though which the dielectric leakage occurs.

You did not get the question, and I think no one did.

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Born2bwire

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The potential difference is the amount of work required to move a charge in an electric field. Inside a conductor, there is no electric field at steady state

The work done in a conductor is by virtue of collision of collisions in on the kernels rather than the charges working against a field...so its not necessary that work by a charge is only done in a field.

Anyway since we're considering superconductors, there will actually be no work done inside the conductor...so if we're connecting the inner side of the plates with superconductor, the the current will be infinite, so to put things in terms of real life cases, lets add a resistance...to this...so I'm asking what will be the energy dissipated through that resistance in an infinitely small time interval.

I'm not getting you...I decoded what you said as work cannot be done by a capacitor.

This is why I think so -

"We can compute stuff using Q = CV, which's a general relation for capacitors.

The Q here will be the net charge accumulated and V will be the potential difference across the plates of the parallel plate air capacitor; note that while using this formula, the arrangement is taken as an 'object' rather than a 2-plate arrangement that is there will be a difference between “the potential difference in each plate by virtue of the charge stored on each of them” and the V that we assume here (that just appears on the back side of the plate).

So...how do we know that there's so much charge on the plates?...how do we measure it and how do we get the 'real' potential “by virtue of the charge stored on each of them”?

There are 2 ways...the simple one is to separate the 2 plate, that way the charge will get exposed throughout the plate, and we'll be able to see its actual potential. On doing so, the V assumed will be different from the potential difference between the plates (that appears on dismantling the arrangement), why...cause on dismantling the plates the capacitance will decrease; remember that the capacitance (or charge per unit volts) of the arrangement and the 2 plates as separates entities are different, the dismantled situation having a lower capacitance...so on dismantling, though the capacitance will decrease, the charge wont...so what will happen?...the P.D will increase to as to store that specific amount of charge on both the plates (a property of their individual capacitance).

The other way is to do so is to connect the plates form the inside, since almost all of the charge has been accumulated on the inner side of the plates, the density will be high, and so will the potential (that's why a dielectric leaks). Connecting the plates from the outer side wont give you that much a potential difference, it will be equal to that of the voltage source.

It cannot be stated that this V computed from between the plates of the capacitor will be equal to the V assumed in Q = CV, that's cause the charge stored on the plate is extremely high and as said before, by virtue of that charge stored the P.D should also be extremely high; the only way the P.D can be reduced is when there'll be a charge difference on both the plates so as to make a lower P.D (equal to V)...but apparently the charge stored on each of the plates is equal and opposite...so this can't be said. Measuring this potential difference (between the 2 plates from the inside) will be like directly measuring the P.D between the 2 plates infinite distance apart (as done before), and that will definitely be higher than the predefined V.

Since the potential difference towards the inner side will be high, the capacitance of the arrangement will be low if that P.D is measured."

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Born2bwire

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This is not true, if that's the case then why is the calculation of the voltage between the plates wrong in the first place? I can't follow your logic here, but I can assure you that you are doing something wrong to come to your conclusion. You should note that when we calculate the voltage and charge distribution of an ideal 1D parallel plate capacitor that we do not bother with terminals or care about the thickness of the plates. The plates can be infinitesimally thin, in which case there isn't a difference between the inner or outer sides, they are the same.It cannot be stated that this V computed from between the plates of the capacitor will be equal to the V assumed in Q = CV, that's cause the charge stored on the plate is extremely high and as said before, by virtue of that charge stored the P.D should also be extremely high

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Born2bwire said:I'm not assuming superconductivity of any sort here. When we talk about conductors in basic electrostatics we are talking about perfect electric conductors. But, if you want to consider a realistic conductor, the answer is the same within normal lab equipment precision. The conductivity of copper and any good conductor is on the order of 10 M Siemens, compare that to sea water which is on the order of 1 Siemen. The resistive lost between the terminal and plate of the conductor will be negligible. We often just replace metals with a perfect electrical conductor because the difference in the results are generally lower than the accuracy of our calculations. Either way, the answer is the same, because there is no electric field residing in a conductor, the voltage drop between the terminal and any point on the inside of the attached plate is zero. Thus, the voltage between the terminals is the same as the voltage between two points on the inside of the opposite plates.

I don't know but to me it seems like a mixup of electric field and conductors...anyway, it doesn't have to do with the topic at hand. :tongue2:

This is not true, if that's the case then why is the calculation of the voltage between the plates wrong in the first place? I can't follow your logic here, but I can assure you that you are doing something wrong to come to your conclusion.

Yeah...that's the problem, I don't know what's wrong, this is the conclusion that I came to using/following basic principles, but this is not seen in real life.

Can you pls quote the things that you're not getting?

You should note that when we calculate the voltage and charge distribution of an ideal 1D parallel plate capacitor that we do not bother with terminals or care about the thickness of the plates.

Thickness doesn't matter anyway.

The plates can be infinitesimally thin, in which case there isn't a difference between the inner or outer sides, they are the same.

Most probably this is a misunderstanding, an infinitely thin plate has the ability to rupture or even terminate a field, it can cause major impact if earthed.

We can't ignore a standing thin, ideal and earthed conductor in a field, it ought to have an impact.

mr.survive said:,V=Q/C..now charge accumulation may vary but have u noticed C ?

C acts as a constant of proportionality for the individual body/capacitor. I've attached a section of my notes to describe what it is...I've considered that too.

why dont u calculate it in simple way where dV=Edr

I'm doing analysis of what is happening, calculating is and figuring out the answer is not the reason, I need an explanation of the phenomenon.

Anyway, in https://www.physicsforums.com/showthread.php?p=2174515#post2174515" thread we discussed that and came to the conclusion that the expression in native form is useless for a capacitor.

n u already know E for II plate capacitor s E=sigma/?

No I don't.

Simply doing a mathematical analysis without having the slightest understanding of the theory is sheer stupidity...this is what is ruining the education system at least in this god damned place.

If only do mathematical analysis, you will most probably end up with a wrong conclusion/answer unless the same sorta question is asked thousand times like in this place; furthermore you will not be able to answer a question out of the usual 'numerical' question.

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It cannot be stated that this V computed from between the plates of the capacitor will be equal to the V assumed in Q = CV, that's cause the charge stored on the plate is extremely high and as said before, by virtue of that charge stored the P.D should also be extremely high

I'll try and make this clearer by adding 3 words -

It cannot be stated that this V computed from between the plates of the capacitor will be equal to the V assumed in Q = CV, that's cause the charge stored on the plate is extremely high and as said before, by virtue of that charge stored the P.D *on each plate* should also be extremely high.

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1) Assume there is a potential difference from one side of a plate to the other.

2) Charge will flow until the diference in potential is zeo.

3) The potential across any plate under static conditions is zero.

4) Therefore the potential difference is the same, measured from the inside of the plates, or the outside of the plates.

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Born2bwire

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I'll try and make this clearer by adding 3 words -

It cannot be stated that this V computed from between the plates of the capacitor will be equal to the V assumed in Q = CV, that's cause the charge stored on the plate is extremely high and as said before, by virtue of that charge stored the P.D *on each plate* should also be extremely high.

Why do you think that the charge has to be high? Take a look at cabraham's derivation from the previous capacitor thread. You can solve for the voltage and charge distribution on the parallel plates. If you want to do it for a finite plate size, there are a myriad of ways to do it by using a method of moments technique, I think Harrington's book has an example of this in his electrostatics section. Either way, I'm not sure why the charge has to be "extremely" high and not be correlated to the potential difference that we solved.

As for the thickness of the plate. I did not say that an infinitesimally thin plate would not have an effect. The capacitor would have the same voltage regardless of the thickness, as you say, and that is my point. With an infinitesimally thin plate, there is no inner or outer side. So the paradox that the inner voltage would not be the same as the outer voltage will not work here because the inner and outer surfaces are the same. However, you will still get the same voltages and charges if you solved assuming an infinitesimally thin plate.

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Phrak said:2) Charge will flow until the diference in potential is zeo.

This point can be proved wrong if you place 2 equally but opposite charge plates in front of each other and earth one...what will happen?...charge will double on the earthed plate, so the difference in potential has been doubled between the plates.

4) Therefore the potential difference is the same, measured from the inside of the plates, or the outside of the plates.

aaaaa...I sorta didn't get any relations between this and the previous points.

Born2bwire said:Why do you think that the charge has to be high?

Ok...what I mean by 'high' is as compared to a single plate, the charge stored on it by virtue of the potential of the charging source will be very low as to when the same plate will be in a capacitor arrangement...so the charge stored on it will be very high, and this charge will get exposed on dismantling the capacitor arrangement.

Take a look at cabraham's derivation from the previous capacitor thread.

I dont know anything about vector calculus

Anyway, mathematical analysis using this logic (that the P.D will be different for the inner and outer side) can also be done.

So the paradox that the inner voltage would not be the same as the outer voltage will not work here because the inner and outer surfaces are the same. However, you will still get the same voltages and charges if you solved assuming an infinitesimally thin plate.

humm...but you still have some thickness...I mean taking it as 0 is not ok...the thickness is extremely less or negligible but not 0...so there will exist 2 faces "infinitely small distance apart".

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Born2bwire

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humm...but you still have some thickness...I mean taking it as 0 is not ok...the thickness is extremely less or negligible but not 0...so there will exist 2 faces "infinitely small distance apart".

No, you can assume that it's thickness is 0. We do this when we do electromagnetic simulations.

aaaaa...I sorta didn't get any relations between this and the previous points.

3) The potential across any plate under static conditions is zero.

4) Therefore the potential difference is the same, measured from the inside of the plates, or the outside of the plates.

This is what I have been repeating through this whole thread. The is no voltage drop across any two points on a continuous conductor when we are at steady state in electrostatics. This is because there is no electric field inside a conductor. This is because charges will always arrange themselves such that they perfectly cancel out any applied electric field in a conductor. Thus, because there is no electric field inside a conductor, then no work against an electric field needs to be done to move a charge which means that the voltage drop is zero. This means that the voltage between the inner and outer edges on a the same plate is zero.

And again, I can see no logic as to your claims. You say previously:

A manifestation of this fact is that, if the arrangement is dismantled, the P.D between the dismantled plates will be much higher than P.D of the charge source.

But there is no explanation to support this. You want to know the charge on the plates? You solve the Laplacian like cabraham showed you. You want to know why the voltage between the inner and outer surfaces of a plate is zero? You solve for the voltage using your knowledge of conductors and electrostatics.

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No, you can assume that it's thickness is 0. We do this when we do electromagnetic simulations.

Without any reason? :surprised

This is what I have been repeating through this whole thread. The is no voltage drop across any two points on a continuous conductor when we are at steady state in electrostatics. This is because there is no electric field inside a conductor. This is because charges will always arrange themselves such that they perfectly cancel out any applied electric field in a conductor. Thus, because there is no electric field inside a conductor, then no work against an electric field needs to be done to move a charge which means that the voltage drop is zero. This means that the voltage between the inner and outer edges on a the same plate is zero.

Oh...now I get it.

But there is no explanation to support this.

:rofl: The whole thing is itself an explanation...how can you explain an explanation :rofl:

Anyway, I think I first have to see to it why a P.D or even a field does not exist in a conductor.

No.

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Me said:2) Charge will flow until the diference in potential is zeo.

This point can be proved wrong if you place 2 equally but opposite charge plates in front of each other and earth one...what will happen?...charge will double on the earthed plate, so the difference in potential has been doubled between the plates.

No, no. This was in reference to a single plate. If there is a potential gradiant inside a conductor, there is an elecric field inside the conductor. Charge will flow under the influence of the electric field. Charge will flow such as to bring the electric field to zero throughout the conductor--the resultant steady-state condition.

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not so. not through these mechanisms.

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