Is There a Perpetual Motion Machine Hidden in the Capacitor Paradox?

[Privalov]

Please see if you can resolve the following relativistic paradox for me.

Capacitor is being charged at rest and accelerated to relativistic velocity.

If we turn capacitor plates perpendicular to the direction of motion, width of dielectric contracts and amount of energy, stored in the capacitor, decreases. See left picture.

If we turn capacitor plates parallel to the direction of motion, length of capacitor contracts, identical charges become closer to each other and energy, stored in the capacitor, increases. See right picture.

If that’s correct, we can discharge capacitor at parallel state and recharge it at perpendicular state. We will get the perpetual motion machine.

 

[Naty1]

I don't believe there is a paradox.

Relativistic length contraction affects any charge as well as the dielectric material...if one expands or contracts so does the other...it's space itself contracting or expanding...so I see no change in energy except for the kinetic energy of motion.

Besides, in the frame of the relativistic capacitor, all remains fixed...there is no contraction unless viewed from another reference frame.

 

[Dale]

You will find all of your missing energy in the magnetic field.

 

[Vanadium 50]

Dale, I'm not sure that's correct.

Suppose in the rest frame, you have an area A and a separation d. If you are moving along the direction of the gap, you now have an area A and a separation d/\gamma for a total volume Ad/\gamma. Now, rotate it 90 degrees. Now you have an area A/\gamma and a separation of d, again for a total volume Ad/\gamma.

Since the energy stored in an electric field is E^2V, since the field is the same and the volume is the same, energy is conserved. No magnetic field is required - at least not to solve this particular problem.

 

[Lok]

As it is already said there is no paradox. relativity works in both ways so...

Imagine someone in a close to c spaceship passing close to you while you hold a capacitor and turn it which way u want. For the spaceship you will appear as to moving with a speed close to c. So the spaceship is the observer now. But you have no paradox in your reference frame as there is no energy difference.

 

[Dale]

a total volume Ad/\gamma. ... since the volume is the same, energy is conserved.

Huh? Ad/\gamma \ne Ad = V.

 

[diazona]

Huh? Ad/\gamma \ne Ad = V.

Actually V = Ad/\gamma, if I understood correctly.

Anyway, Vanadium 50 was really just saying Ad/\gamma = Ad/\gamma. Can't argue with that...

 

[Dale]

Ahh, I was answering a different question! I understand Vanadium50's point now. He is absolutely correct. There is a magnetic field also in the moving frame, and like the electric field in the moving frame it has the same volume and same energy density in both the parallel and perpendicular orientations.

 

[Bob S]

As long as the electric field in the capacitor is parallel to the direction of motion, the changes are limited to the relativistic contraction of the capacitor dimensions along the direction of motion. When the capacitor is rotated by 90 degrees, so that the electric field is perpendicular to the direction of motion, two additional things happen: See http://pdg.lbl.gov/2004/reviews/elecrelarpp.pdf
1) The component of E perpendicular to the motion, E_p, is increased by a factor γ, so that
E_p' = γE_p where the prime indicates the field observed in the system moving relative to capacitor.

In addition, the electric field perpendicular to the direction of motion creates a magnetic field (as suggested by daleSpam) B_p' given by
B_p' = -γ(1/c²)(v x E_p)
So the integral of the stored energy over the volume now becomes

W = (1/2) ∫(E_p' ² + B_p' ²) dV' = (1/2) ∫ (γ²E_p² + (γ²β²/c⁴)(c x E_p)²) dV' (where dV is a volume element)

[Edit] added μ₀ and ε₀
W = (1/2) ∫(ε₀E_p' ² + B_p' ²/μ₀) dV' = (1/2) ∫ (γ²ε₀E_p² + (γ²β²/c⁴)(c x E_p)²/μ₀) dV' (where dV is a volume element)
So
W = (1/2) ∫ (γ²ε₀E_p² + (γ²β²/c²)E_p²/μ₀) dV'

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
∂ C ∏ ∑

 

[Privalov]

You will find all of your missing energy in the magnetic field.

I’m not missing energy; I’m getting infinite amount of it.

Since the energy stored in an electric field is E^2V

Wikipedia seems to give slightly different formula:

http://en.wikipedia.org/wiki/Electric_field#Energy_in_the_electric_field

Anyhow, this is irrelevant minor correction.

since the field is the same and the volume is the same, energy is conserved.

Volume is constant; however, electric field is not constant in this problem.

Electric field between (indefinitely large) capacitor plates is given by equation

E = Q / A \epsilon

Where
E is electric field;
Q is the charge (I believe it’s invariant and should be the same in all frames of reference);
A is surface area of the plates;
\epsilon is permittivity.

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html

Surface area is subject to relativistic length contraction. Area depends on orientation of the capacitor. Therefore, stored energy can increase, when someone turns the capacitor, without any energy intake from external sources. I see a paradox here.

 

[meopemuk]

Hi Privalov,

for the last 100 years your paradox has been known as the Trouton-Noble paradox. If you search literature, you can find many attempts to explain it from the point of view of classical electrodynamics. All these attempts look rather unconvincing.

 

[Phrak]

Hi Privalov,

for the last 100 years your paradox has been known as the Trouton-Noble paradox. If you search literature, you can find many attempts to explain it from the point of view of classical electrodynamics. All these attempts look rather unconvincing.

This would be shocking. Do you have a reference?

 

[Naty1]

Wikipedia describes this:

http://en.wikipedia.org/wiki/Trouton-Noble_experiment

The Trouton–Noble experiment attempted to detect motion of the Earth through the luminiferous aether, and was conducted in 1901–1903 by Frederick Thomas Trouton (who also developed the Trouton's ratio) and H. R. Noble. It was based on a suggestion by George FitzGerald that a charged parallel-plate capacitor moving through the aether should orient itself perpendicular to the motion.

Sounds like a different effort?

 

[Naty1]

Privalov posted:

Electric field between (indefinitely large) capacitor plates is given by equation

Where
is electric field;
is the charge (I believe it’s invariant and should be the same in all frames of reference);
is surface area of the plates;
is permittivity.

http://www.ac.wwu.edu/~vawter/Physic...ParallCap.html

Surface area is subject to relativistic length contraction. Area depends on orientation of the capacitor. Therefore, stored energy can increase, when someone turns the capacitor, without any energy intake from external sources. I see a paradox here.

[So why would this frame reference issue by any different than,say, length contraction...what you see depends on your frame??]

The above is basically what I was thinking in my post #2; but I now wonder if infinite plates are a suitable model since they will be infinite in any frame.

Here is what Peter Bergmann, a student of Einstein, says in his 1992 book THE RIDDLE OF GRAVITATION, page 45 which I believe relates. The overall description is quite complex:

..electric charge of a physical system is a scalar whose value is the same in all coordinate systems...the electric charge of an isolated physical system is conserved... charge density (reflects) different volumes in different Lorentz frames...the charge density is not the same for all observers..

It goes on for another four or five long paragraphs regarding four dimensional vectors and I can't get the gist of it...I am not comfortable trying to summarize it...If anyone has access to the book and knows the math references, I believe it applies to this thread issue.

 

[Naty1]

And while we are at it: isn't there an electric field in the frame of the capacitor, but no magnetic field, while the moving observer sees both E & M?

Also when charges are in relative motion, and I'm not sure if the observer in motion would observe such motion or not, there would be a current density, a flux. Is that observed or not?? And how might it relate to the OP questions??

 

[Dale]

There is clearly a magnetic field for the moving observer since the moving charge is, by definition, a current. Since http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" for Maxwell's equations the details of the scenario are irrelevant. If you get that energy is not conserved then you have necessarily violated one or more of Maxwell's laws. I think that it is obvious that the OP's analysis violates Maxwell's equations since the energy in the magnetic field is not even considered, but I leave it up to the OP or other interested readers to derive the details.

 

[Privalov]

http://en.wikipedia.org/wiki/Trouton-Noble_experiment
Sounds like a different effort?

Sounds like a different effort to me.

 

[Bob S]

There is clearly a magnetic field for the moving observer since the moving charge is, by definition, a current. Since http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" for Maxwell's equations the details of the scenario are irrelevant. If you get that energy is not conserved then you have necessarily violated one or more of Maxwell's laws. I think that it is obvious that the OP's analysis violates Maxwell's equations since the energy in the magnetic field is not even considered, but I leave it up to the OP or other interested readers to derive the details.

In my post above, I found that a) there is a magnetic field for the moving observer, and b) the stored energy may not be conserved. Specifically
I showed
W = (1/2) ∫ (γ²ε₀E_p² + (γ²β²/c²)E_p²/μ₀) dV'
which becomes, when substituting ε₀μ₀=1/c²

W = (1/2)γ²ε₀∫(1+β²)E_p² dV'
The real problem is whether there should be a + sign in (1+β²)? It arises from summing the stored energy from electric (E²) and magnetic (B²) fields. To make the stored energy independent of rotation, it should be a - sign.


α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

 

[Dale]

There must be a mistake then. Energy is provably conserved in general as I linked to above, so the specific situation does not matter. Also, the total energy is the sum of the energy in the electric and magnetic fields. That is what was proven to be conserved.

 

[meopemuk]

Sounds like a different effort to me.

According to your assumption, the energy of a moving capacitor depends on the orientation of its plates. This means that there should be a torque which tends to rotate the moving capacitor to a preferable orientation with respect to the movement direction, so that the energy is lowered. This is exactly the Trouton-Noble idea.

Here are some references:

F. T. Trouton and H. R. Noble, "The Mechanical Forces Acting on a Charged Electric Condenser Moving through Space", Phil. Trans. Roy. Soc. London A, 202 (1904), 165.

G. Spavieri and G. T. Gillies, "Fundamental tests of electrodynamic theories: Conceptual investigations of the Trouton-Noble and hidden momentum effects",
Nuovo Cim., 118B (2003), 205.

J. Franklin, "The lack of rotation in the Trouton-Noble experiment", http://www.arxiv.org/abs/physics/0603110v3

S. A. Teukolsky, "The explanation of the Trouton-Noble experiment revisited", Am. J. Phys., 64 (1996), 1104}

O. D. Jefimenko, "The Trouton-Noble paradox", J. Phys. A: Math. Gen., 32 (1999), 3755.

J. D. Jackson, "Torque or no torque? Simple charged particle motion observed in different inertial frames", Am. J. Phys., 72 (2004), 1484.

 

[Vanadium 50]

Let's try and simplify this a bit.

First, this is a problem about rotations, not boosts - it's what happens to an already boosted system when it's rotated. That means we don't need to solve this in the most general case, and as we saw, we don't have to worry about electric and magnetic fields changing identities.

Second, the answer to this will follow from Trouton-Noble (it's a special case) but as Trouton-Noble is quite confusing (as evidenced by people still writing about it a century later), I don't think this is a particularly enlightening approach. I suggest we discard it.

My answer in #4 has only a few steps. Privalov, it would help if you would indicate which, if any of the following statements you disagree with:

(1) If the energy stored in the electric field in a parallel plate capacitor remains unchanged, the energy stored in the capacitor remains unchanged.

(2) The electric field between the plates does not depend on the orientation of the capacitor.

(3) The volume between the plates does not depend on the orientation of the capacitor.

(4) Given (1), (2) and (3), one can show the energy in the capacitor does not depend on it's orientation.

 

[Bob S]

(1) If the energy stored in the electric field in a parallel plate capacitor remains unchanged, the energy stored in the capacitor remains unchanged.
(2) The electric field between the plates does not depend on the orientation of the capacitor.
(3) The volume between the plates does not depend on the orientation of the capacitor.
(4) Given (1), (2) and (3), one can show the energy in the capacitor does not depend on it's orientation.

The relativistic transformation of the electric field in the capacitor creates a magnetic field if the electric field is orthogonal to the direction of motion, but not if it is parallel. See
http://pdg.lbl.gov/2004/reviews/elecrelarpp.pdf
The total energy in the capacitor should not depend on orientation. The volume does not change, so the energy density should not change either. But if the magnetic and electric field densities are summed, it does. However, if the densities are subtracted, it does not. Here is the total energy (see my posts # 9 and # 18)

W = (1/2)γ²ε₀∫(1+β²)E_p² dV'

Why is it (1+β²) and not (1-β²) in the above integral? If it were the latter, the energy density would be independent of orientation.

 

[Privalov]

According to your assumption, the energy of a moving capacitor depends on the orientation of its plates. This means that there should be a torque which tends to rotate the moving capacitor to a preferable orientation with respect to the movement direction, so that the energy is lowered. This is exactly the Trouton-Noble idea.

I believe I came up with much more paradoxical conclusion, than undetectably minor torque: the violation of conservation laws. So, I'm trying to understand, what is wrong with my logic. I'm not trying to find luminiferous aether.

(1) If the energy stored in the electric field in a parallel plate capacitor remains unchanged, the energy stored in the capacitor remains unchanged.

True.

(2) The electric field between the plates does not depend on the orientation of the capacitor.

True in classical physics; false in relativity.

(3) The volume between the plates does not depend on the orientation of the capacitor.

True in both classical physics and relativity.

Given (1), (2) and (3), one can show the energy in the capacitor does not depend on it's orientation.

The only problem is (2) is false.

As long as the electric field in the capacitor is parallel to the direction of motion, the changes are limited to the relativistic contraction of the capacitor dimensions along the direction of motion.

The paradox exists, based on that statement alone.

Energy in any form or storage, moving with velocity v, must increase by a factor of y, if measured by observer at rest. However, relativistic contraction of distance between capacitor plates makes the amount of energy, stored in capacitor, smaller, by a factor of y.

By the way, even if you will be able to write electromagnetism equations and show there is no paradox there, it won’t really resolve the original paradox.

Please note original paradox is formulated in terms of special relativity, not electromagnetism. However, electromagnetism can be derived from special relativity, if we take into account relativistic contraction of distance between moving changes and their time dilation. Therefore, if irresolvable paradox will be found in special relativity, but not in electromagnetism, that will be a paradox itself. Since one is derived from the other, its conclusions must be identical.

However, resolution of this paradox from electromagnetism perspective might really help.

 

[Phrak]

The relativistic transformation of the electric field in the capacitor creates a magnetic field if the electric field is orthogonal to the direction of motion, but not if it is parallel. See
http://pdg.lbl.gov/2004/reviews/elecrelarpp.pdf
The total energy in the capacitor should not depend on orientation. The volume does not change, so the energy density should not change either. But if the magnetic and electric field densities are summed, it does. However, if the densities are subtracted, it does not. Here is the total energy (see my posts # 9 and # 18)

W = (1/2)γ²ε₀∫(1+β²)E_p² dV'

Why is it (1+β²) and not (1-β²) in the above integral? If it were the latter, the energy density would be independent of orientation.

Thanks for bringing this up Bob. The energy, E²+B² has always bothered me because it's not an invariant quantity, and thus suspect as a useful quantity.

You're equation would be correct if you subtracted--if you begin with E²-B². You've described a Lorentz boost of an electric field perpendicular to the boost direction. Of course, energy in general is not conserved under a Lorentz transformation, and neither is the electromagnetic field energy, nor the field energy density in general.

E²-B² is the conserved quantity under a Lorentz transform (where c=1), and a general linear transform, as well (which is really nice to find out, because, although total energy is not converved in curved spacetime, E²-B² is).

 

[Bob S]

Thanks for bringing this up Bob. The energy, E²+B² has always bothered me because it's not an invariant quantity, and thus suspect as a useful quantity.
E²-B² is the conserved quantity under a Lorentz transform. (where c=1) and a general linear transform, as well.

Thanks. My equation shows that the stored energy is rotation invarient if E²-B² is the conserved quantity..
Bob S

 

[Phrak]

Thanks. My equation shows that the stored energy is rotation invarient if E²-B² is the conserved quantity..
Bob S

Can you post that? I haven't worked it out.

 

[Vanadium 50]

Privalov, it would help if you would indicate which, if any of the following statements you disagree with:

(2) The electric field between the plates does not depend on the orientation of the capacitor..

True in classical physics; false in relativity.

Okay, now we're making some progress. (By the way, I clearly should have written "magnitude of the electric field" above, but it's equally clear that you understood it.)

As I mentioned before, this problem does not involve boosts. It involves rotations of a system already boosted. Even in special relativity, rotations do not mix components of E and B: they mix E_x with E_y and E_z, and B_x with B_y and B_z, but not E and B. (Boosts do that) The magnitude of E and B however, remains invariant.

 

[Phrak]

As I mentioned before, this problem does not involve boosts. It involves rotations of a system already boosted.

But it does involve a boost. One is boost. The other is boost then rotate, or rotate then boost. This is not equivalent to rotate only.

 

[Vanadium 50]

I think that's exactly where the problem lies. The problem as described is the rotation of an already boosted system. People are running into difficulties because they want to analyze this without properly factorizing the problem in such a way that it looks simple.

 

[Dale]

To Bob S and Phrak,

In general a conserved quantity is not necessarily invariant, and an invariant quantity is not necessarily conserved. They are completely different ideas. The energy conservation law in EM is:

\frac{\partial U}{\partial t} + \nabla\cdot{\bf u} = -{\bf E}\cdot {\bf j}
where
\frac{\epsilon_0 E^2}{2} + \frac{B^2}{2 \mu_0}
and
{\bf u} = \frac{{\bf E}\times{\bf B}}{\mu_0}

(see http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" )

If you get that it is not conserved then you have necessarily violated Maxwell's laws. By the way, my guess for the specific way that Maxwell is violated is that the approximation of a uniform E-field only applies for an infinite sheet of charge, whose length cannot contract (being infinite). When determining the energy of a realistic capacitor the E-field is not uniform nor contained between the plates. In fact, a dipole is probably a better approximation.

 

[Dale]

I believe I came up with much more paradoxical conclusion, than undetectably minor torque: the violation of conservation laws. So, I'm trying to understand, what is wrong with my logic. ... resolution of this paradox from electromagnetism perspective might really help.

Again see:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html
What is wrong is simply that you violated Maxwell's equations in your analysis. It is the only possible way to get a violation of the conservation of energy since it stems directly from Maxwell's laws. For your analysis the biggest violation is obvious: you ignored the magnetic field. But there are probably other errors as well as I mentioned in the above post to Bob S and Phrak.

 

[Naty1]

This is a fascinating thread! Love it! still confusing me, however...

Dalespam posted:

There is clearly a magnetic field for the moving observer since the moving charge is, by definition, a current.

Yes. But am I'm trying to understand a related issue. What does "moving" here mean...usually we mean a charge undergoes a displacement; that's a current flow, a flux.

But here the space is moving the charge along with it. The thing that still confuses me is if length contraction is observed from the moving reference, is there a current flow?

The charge contracts by riding along the space contraction simultaneously as space contracts ...there is no movement of charge relative to observed contracting space ...does this constitute an observed flux??

In other words, there are two ways to increase charge density: move more charge into a fixed volume (clearly a case of current flow) or compress a volume carrying a fixed quantity of charge into a smaller volume... Are these precisely equivalent in relativity??

I'm now thinking "yes".

 

[Dale]

The http://en.wikipedia.org/wiki/Four-current" (more correctly the four-current-density) transforms as a four-vector. It's components are (cρ,j) where ρ is the ordinary charge density and j is the ordinary current density. So if you have a static charge density in one frame (cρ,0) then you will get (γcρ,γvρ) in any other frame. So, current is indeed flowing in the other reference frame i.e. because it is a component of the four-current.

 

[meopemuk]

I believe I came up with much more paradoxical conclusion, than undetectably minor torque: the violation of conservation laws.

Yes, I agree that your idea is somewhat different from the Trouton-Noble paradox. However, note also some similarities. In both cases we are dealing with conservation laws. The Trouton-Noble idea concerns the conservation of the total angular momentum (this conservation law requires zero torque on the moving capacitor). You are discussing the conservation of energy.

 

[Privalov]

But it does involve a boost. One is boost. The other is boost then rotate, or rotate then boost. This is not equivalent to rotate only.

You can think about it as a boost and multiple rotations.

To make the stored energy independent of rotation, it should be a - sign.

Is it possible the equations you have derived are the electromagnetic representation of the same paradox? You came up with the same “margin of error”, as I did.

 

[Phrak]

The http://en.wikipedia.org/wiki/Four-current" (more correctly the four-current-density) transforms as a four-vector.

Now that you've brought it up, could you clear up a point for me? I hope you don't mind the word doc attachment; the equations were difficult enough as was.

 

[Dale]

My understanding was that assigning units to the tensor itself was the same as assigning units to each of the components. With the caviat that the components are scalars so a tensor with units X cannot be added to a component with the same units. That could be wrong, but it was my understanding.

 

[Phrak]

My understanding was that assigning units to the tensor itself was the same as assigning units to each of the components. With the caviat that the components are scalars so a tensor with units X cannot be added to a component with the same units. That could be wrong, but it was my understanding.

If I have a space-time vector (type 1,0 tensor) whose elements have units (T,D,D,D) and attach units of per-volume, V^-3, to the entire vector I get units (T,D,D,D)V^-3, or (TV^-3, DV^-3, DV^-3, DV^-3).

I don't know if that's quit what you said, but it makes sense that a type(1,0) tensor can have elements with charge densities now you've talked it up. Thanks for helping me over this thing. It's been bothering me for better than a year! And now seems perfectly obvious.

 

[Dale]

Well, usually we multiply the T by c to get units of T L/T = L. This is called making it dimensionally consistent and is really required if you are going to do any summation over the elements. So the standard four-vector has units (T L/T, L, L, L) = (L, L, L, L) = (1,1,1,1) L.

In the case of the four-current, ordinary current density is current/area and current is charge/time, so current density is C/TL². The charge density on the other hand is charge/volume C/L³ which when multiplied by c gives C/L³ L/T = C/TL² which is the same units as current density. So the four-current is also dimensionally consistent with units of current/area.

 

[Naty1]

And so can someone summarize what has been agreed to here and why?? OR what has not been agreed to and why??

It sure would be nice if a Science Advisor would briefly summarize conclusions in long threads like this. Or identify points of remaining contention.

That would not only aid those of us who have not formally studied all the underlying math and theory, but might reduce the number of repeat questions on the same issue...

 

[Dale]

Yeah, but the summary would inevitably wind up being as disputed as the rest of the thread. E.g. I would summarize this thread as follows:

Energy is conserved in Maxwell's equations. You can, of course, ignore Maxwell's equations and then mistakenly conclude that energy is not conserved. One specific example of how to do so is the OP.

Let the objections commence.

 

[DrGreg]

Well, usually we multiply the T by c to get units of T L/T = L. This is called making it dimensionally consistent and is really required if you are going to do any summation over the elements. So the standard four-vector has units (T L/T, L, L, L) = (L, L, L, L) = (1,1,1,1) L.

In the case of the four-current, ordinary current density is current/area and current is charge/time, so current density is C/TL². The charge density on the other hand is charge/volume C/L³ which when multiplied by c gives C/L³ L/T = C/TL² which is the same units as current density. So the four-current is also dimensionally consistent with units of current/area.

What DaleSpam says is true, but note the word "usually". I think I would prefer to say "...often, in special relativity,..."

It's not compulsory to work with "dimensionally consistent" components, indeed in general relativity we often don't. E.g. two of the components could be angles, not lengths, in spherical polar spatial coordinates. The issue of summation over the elements gets handled by the components of the metric tensor in your chosen coordinates.

As an aside, I only recently realized that in (T, L, L, L) Minkowski coordinates the covariant (i.e. not contravariant) 4-momentum comes out in (E, p, p, p) units without any conversion factor required.

Which was nice.

 

[Dale]

Yes, I agree with DrGreg, my comments were the usual convention for special relativity, not general relativity which is much more flexible in that sense and doesn't require dimensional consistency of tensor elements in the same way.

Thanks for clarifying DrGreg.

 

[Privalov]

My equation shows that the stored energy is rotation invarient if E²-B² is the conserved quantity..

Energy in electric field is the potential energy of separated charges. It can change, if the distance between charges changes. In principle, it could have changed in the given problem (it did not though).

Energy in magnetic field is the kinetic energy of moving charges. In this paradox, speed of charges remains the same. Therefore, magnetic field energy remains constant. I believe electromagnetism has nothing to do this particular problem.

 

[Dale]

Energy in magnetic field is the kinetic energy of moving charges.

This is not correct at all. The KE of charges is negligible, typically the electron drift velocity is on the order of 1 mm/s or even less. With so little velocity and so little mass their KE is typically insignificant and in general has nothing to do with the energy of the magnetic field.

 

[Dadface]

I have read through this several times trying to get the gist of it but I am still confused so I apologise in advance if I get it wrong and misinterpret some of which has been written.
Firstly the energy density is given by n=0.5* permittivity*E^2
Secondly it seems to have been suggested that since the volume is the same in each case then so is the energy.This is not so because the capacitance of the capacitor is not proportional to its volume it is proportional to the plate overlap area(A) divided by the plate separation(d).Here I have considered a parallel plate capacitor and I have ignored edge effects.It follows that if A reduced to half of its original value then so would C reduce this resulting in a doubling of the energy stored(Energy=Q^2/2C).I have assumed that the charge (Q) remains constant.If d is reduced to half of its original value C doubles and the energy stored halves.

 

[pallidin]

Take 2 pennies and separate them equal to the thickness to one penny. The air is the dielectric.
Charge those pennies like a capacitor, and note their discharge potential.

Now, take those two pennies and beat them down to such a thickness that it is "paper-thin", such as each penny is now about 6-inches in diameter.

With the same dielectric(air) and dielectric spacing(one penny thickness), charge and note discharge potential.

The second scenario has much greater energy storage potential.

 

[Privalov]

The KE of charges is negligible, typically the electron drift velocity is on the order of 1 mm/s or even less. With so little velocity and so little mass their KE is typically insignificant

Yes; and energy in magnetic field of a typical wire at any given moment is also insignificant. However, any attempt to drain the energy from this magnetic field will result in more energy supplied by the power plant. So, total amount of energy, transferred by wire, can become significant. Kinetic energy of moving electrons remains the energy carrier though.

Anyhow, this seems like terminological dispute. Let me find some material examples.

I believe that accurate calculation will reveal the energy in magnetic field of a moving capacitor will not depend on capacitor orientation (so capacitor experiences no torque). True or false?

It will be hard to prove though (at least based on formulas, which Bob S used).

Let’s say we have a charged particle moving at given velocity. By placing solenoids around it, we can extract energy from its magnetic field. I believe we can not extract more energy, than the kinetic energy of particle. True or false?

it seems to have been suggested that since the volume is the same in each case then so is the energy.This is not so

Not so, if the voltage is the same. Typically, it’s the case. However, in this particular problem (as described in my post #1) there are no constraints on voltage; while charge is constant.

 

[Dadface]

Take 2 pennies and separate them equal to the thickness to one penny. The air is the dielectric.
Charge those pennies like a capacitor, and note their discharge potential.

Now, take those two pennies and beat them down to such a thickness that it is "paper-thin", such as each penny is now about 6-inches in diameter.

With the same dielectric(air) and dielectric spacing(one penny thickness), charge and note discharge potential.

The second scenario has much greater energy storage potential.

So basically you have increased the capacitance by increasing the plate area.It is true that if you take a large and a small capacitor and charge them to the same voltage then the large one will store more energy.With this paradox we can have different situations:
1.BATTERY PERMANENTLY CONNECTED.In this case V will equalise when the capacitance is varied but Q will vary.If C is increased, more charge will flow from the battery and the energy stored will increase.
2.CAPACITOR CHARGED AND BATTERY DISCONNECTED.In this case Q will be constant but V will vary.If C is increased, V will decrease and the energy stored will decrease.
In my post above I was referring to situation two and although both situations give opposite results the paradox still exists.

 

[pallidin]

Perhaps I'm not understanding. What, specifically, is the paradox?