Energy Conservation - Static Field
ray
accelerating electrons between plates with a static charge.
A simple case for consideration is to have two grids with a static
electric charge such that they have a potential difference. When we
let a group of electrons into the intervening space, they accelerate
and pass through the positive grid. The electrons now have additional
energy due to their acceleration and EM disturbance has propagated
from them while accelerating. But the charged plates are still
charged the same as now current has moved from one to the other.
Where did the energy come from?
Ray
Tom Roberts
TYPO: "now" => "no" ----^ (and you really mean charge, not current)
> Where did the energy come from?
There actually is no "extra" energy; indeed when the electrons are far
from the grids, they will be traveling SLOWER than when they originally
entered the region between the grids. There will also be electromagnetic
radiation traveling away from the region of the grids, generated by the
electrons as they accelerated.
Yes, between the grids the electrons will be attracted to the positive
grid and will bend toward it (you assumed the attraction is strong
enough so the electrons bend sharply enough to go through the positive
grid). But after passing through it, they are still attracted to the
positive grid more than being repulsed by the negative grid, and will be
slowed down by this attraction. In the absence of radiation, far away
from the grids the electrons would have the same kinetic energy (and
therefore the same speed) as they had initially (far before coming
between the grids). But accelerating electrons does generate radiation,
so their final speed must be less than their initial speed. The energy
of the radiation is exactly matched by the loss in kinetic energy of the
electrons.
But energy must be conserved at every instant in time, not just when
they are far away. So where does the energy come from when the electrons
are moving the fastest, right when they pass through the positive grid?
The answer is that this energy comes from the electric field -- clearly
with negative charges co-located with the positive grid, the total E
field from grid+electrons is less than from the grid alone. The reduced
E field implies lower energy density, and the reduction in the total
energy of the field equals the increased kinetic energy of the
electrons. Note the total energy in the field includes the radiation.
Tom Roberts
ray
I am sorry but I don't understand. Once the electrons are in the
field, they will accelerate towards the positive grid. This
acceleration is an increase in the electron's kentic energy. Where
did this energy come from? As it passes through the positive grid, it
will be at some increased speed. Once it is outside the plates, there
is no electric field except for some leakage around the edges, so
there is nothing to slow it down. So it exits at a higer speed than
when it went in. Where did it get the energy?
Ray
Anti Vigilante
Energy is what fields do. Although I think Tom's explanation is quite
clear.
> As it passes through the positive grid, it will be at some
> increased speed. Once it is outside the plates...
If it doesn't get trapped.
Once it is past the positive grid it it will be attracted back to the
positive grid.
>
> Ray
--
Fuck the Enlightenment! Viva la Renaissance!
Tom Roberts
> Tom Roberts wrote:
>> [...] So where does the energy come from when the electrons
>> are moving the fastest, right when they pass through the positive grid?
>> The answer is that this energy comes from the electric field -- clearly
>> with negative charges co-located with the positive grid, the total E
>> field from grid+electrons is less than from the grid alone. The reduced
>> E field implies lower energy density, and the reduction in the total
>> energy of the field equals the increased kinetic energy of the
>> electrons. Note the total energy in the field includes the radiation.
>
> field, they will accelerate towards the positive grid. This
> acceleration is an increase in the electron's kentic energy.
Yes (kinetic energy).
> Where
> did this energy come from?
As I said above, it comes from reduced total energy in the electric field.
> As it passes through the positive grid, it
> will be at some increased speed.
Yes.
> Once it is outside the plates, there
> is no electric field except for some leakage around the edges, so
> there is nothing to slow it down.
Not true. Even if the electron emerges from the center of the grid,
there is still nonzero electric field there pulling it back toward the
grid. Remember that it is closer to the positive grid than it is to the
negative grid, and that the electrostatic force falls inversely as
distance squared. For the unphysical case of infinite (plane and
parallel) grids, the forces from the two grids cancel to zero everywhere
outside them, but for finite grids the forces never cancel.
We know from electrostatic theory that if radiation can be neglected
then electrostatic fields are conservative -- for this case that means
that far away from the grids, the kinetic energy of the electron is the
same before and after getting deflected by them (imagine firing it from
far away into the region between them).
> So it exits at a higer speed than
> when it went in. Where did it get the energy?
As I said, from reduced energy density in the electrostatic field.
Once it gets far away, the situation of the fields near the grids is the
same as before the electron approached, so the total energy in the
fields must be the same; this implies that the kinetic energy of the
electron must also be the same as before it approached them. That's how
we know that it slows back down outside the grids.
If you want to consider infinite grids, then note there
is no way to "fire the electron from far away"; nor can
the electron ever get "far away". There is also infinite
total energy in the field, so one cannot determine
whether it decreases as the electron is accelerated....
Tom Roberts
John Polasek
wrote:
Here is another way to look at it.
Let the electron density be rho, with an active area A. When the
velocity of the cloud is V at surface A, the effect is a current
through A, having the value
I = rho*A*V coulombs per second (amperes)
which is likewise seen in the battery so the power in the beam must be
Pwr = Vbat*I
and if for a period of time T, the energy is
En = Vbat*I*T = Vbat*Q
where Q is the number of coulombs passed.
John Polasek
ray
> On Sat, 5 Dec 2009 09:25:03 -0800 (PST), ray <Ray.Jos...@CDICorp.com>
John,
Thank you for the view. That is a way to look at it but it does not
resolve the issue.
Before the electrons under test are placed in the system, the plates
have a specific charge and geometry and thus a fixed voltage and
energy content. When we release the electrons in the intervening
space, they accelerate through the space and exit the system.
Moments later, the electrons are screaming away, with next to no
external field, these electrons are forever gone into space. But the
plates still have the same charge and the same geometry, thus they
have the same voltage and energy.
We can say that 'things' will happen because they must, but to make it
more than superstition, we must be able to understand why.
Ray
ray
> ray wrote:
> > Tom Roberts wrote:
> >> [...] So where does the energy come from when the electrons
> >> are moving the fastest, right when they pass through the positive grid?
> >> The answer is that this energy comes from the electric field -- clearly
> >> with negative charges co-located with the positive grid, the total E
> >> field from grid+electrons is less than from the grid alone. The reduced
> >> E field implies lower energy density, and the reduction in the total
> >> energy of the field equals the increased kinetic energy of the
> >> electrons. Note the total energy in the field includes the radiation.
>
> > I am sorry but I don't understand. Once the electrons are in the
> > field, they will accelerate towards the positive grid. This
> > acceleration is an increase in the electron's kentic energy.
>
> Yes (kinetic energy).
>
> > Where
> > did this energy come from?
>
> As I said above, it comes from reduced total energy in the electric field.
I am sorry, but I don't see where the electric field gave up any
energy. The plates have a fixed charge on them and a fixed geometry
thus the voltage and energy is still the same.
>
> > As it passes through the positive grid, it
> > will be at some increased speed.
>
> Yes.
>
> > Once it is outside the plates, there
> > is no electric field except for some leakage around the edges, so
> > there is nothing to slow it down.
>
> Not true. Even if the electron emerges from the center of the grid,
> there is still nonzero electric field there pulling it back toward the
> grid. Remember that it is closer to the positive grid than it is to the
> negative grid, and that the electrostatic force falls inversely as
> distance squared. For the unphysical case of infinite (plane and
> parallel) grids, the forces from the two grids cancel to zero everywhere
> outside them, but for finite grids the forces never cancel.
As the electrons emerge from the system, they will see only fringe
field from the plates. Thus the full effect the electrons experienced
inside the system will not be seen out side. TO further express, the
field between the plates is constant through out the space between the
plates, once out side the plates, there is only fringe. As the
particle gets further away, the relative distance between the
electrons are from each plat becomes equal and thus minimal effect.
>
> We know from electrostatic theory that if radiation can be neglected
> then electrostatic fields are conservative -- for this case that means
> that far away from the grids, the kinetic energy of the electron is the
> same before and after getting deflected by them (imagine firing it from
> far away into the region between them).
You are correct in your attempt to invoke conservation but we can not
just assume the ensuing results, we must be able to explain how. For
example, if the high velocity emergent electrons slam into an isolated
target, they don't get to return their energy, the energy they gained
will be transfered to the external target.
Further, as you mentioned, conservation must exist at all tijmes.
When the target electrons emerge from the system, they have a higher
energy than when they went in. At that moment, the energy with the
plates must be the different than when the electrons entered the
system, but there has been no change in the energy parameters of the
plates.
>
> > So it exits at a higer speed than
> > when it went in. Where did it get the energy?
>
> As I said, from reduced energy density in the electrostatic field.
We need to be able to show how the energy density of the field has
changed. The charge on the plates has not changed nor has the
geometry. How did the energy change? To not invoke superstition, we
need to be able to explain how the energy has changed.
>
> Once it gets far away, the situation of the fields near the grids is the
> same as before the electron approached, so the total energy in the
> fields must be the same; this implies that the kinetic energy of the
> electron must also be the same as before it approached them. That's how
> we know that it slows back down outside the grids.
>
> If you want to consider infinite grids, then note there
> is no way to "fire the electron from far away"; nor can
> the electron ever get "far away". There is also infinite
> total energy in the field, so one cannot determine
> whether it decreases as the electron is accelerated....
>
> Tom Roberts
Regards,
Ray
Anti Vigilante
> The plates have a fixed charge on them and a fixed geometry thus the
> voltage and energy is still the same.
>
>
It's really simple:
First of all, the field is not an object. We cannot see it anymore than
we could see photons on their way to their target. We only see what we
receive. It's a natural perception limit that only education remedy.
The field is a mathematical tool. It does not say what goes on to cause
its features. It just says what the outcome will be.
When you learned your times tables you were using a field.
The field therefore is the second step in a three step process. The
distance is merely a limiting factor. It's not so much part of the field
as it is part of what goes on behind the man behind the curtain. It's not
the motivator.
The goal is to discover how an electron will move.
The first step is finding the charge. Naturally that is provided to us.
The second step is to do the calculation for the field at some point.
The third step is to discover the effect on another charged particle at
that point.
We cannot describe changes in motion without acceleration. The second
derivative is necessary. Velocity had to become what it is by
acceleration. Before you can have change, you must have change in the
rate of change. Before you can have the egg you need the egg which
hatched the chicken which laid the egg.
Without acceleration there is no concept of velocity. We learn reality
backwards. Acceleration came first. Then velocity. It's just that we meet
position first before we meet velocity. And acceleration we ignorantly
almost never meet unless we trip and fall. But it is acceleration which
describes reality.
So the force is defined by the second charged partner's charge and the
field. And it is the total field which is the SUM of fields which
determines the force.
The field doesn't give up anything. The field is a tool. However whatever
activity goes on there does give up energy. Why? Because that's why the
field was developed as a tool in the first place.
So how does this exchange work? The shenanigans that make the field give
up something and it just so happens to add/multiply up to the same
numbers as the change in kinetic energy.
Why? BECAUSE THAT'S WHAT IT WAS INVENTED TO DO. That is the whole point
of its creation. The only reason the field is real is because it is
practical. The only reason it is practical is because we use it
consistently and we follow rigid rules in its application. It is the
framework of its intent and the uncompromising abiding by the rules that
makes the field a real tool. But that's all it is.
So here's the numbers:
Efield = q/r^2
A positive charge has a positive field. Pick any point X, Y, Z in the
vicinity and you get a field value. Not a force. A field value.
A negative charge has a negative field.
When a negative charge approaches (x, y, z) the field value decreases.
That decrease is by the proper calculation exactly the same as the change
in the kinetic energy of the negative charge. The grid being more massive
tends not to change as much.
And so the field seems to give up energy to the electron.
But you see this is misleading because the analogy is entirely by luck.
The field value which is reduced tells us the force on a THIRD particle
not the electron. That is that field value corresponds to the
contribution made by both the visiting charge and the grid.
So it is true that the field doesn't give anything up. However it is also
true that the gears and guts under this very elegant field do give up
something. And that is where the energy comes from.
The the total field
ray
wrote:
Anti,
That was a fun ride! Thank you.
I would like to address my original posting where I never mentioned
field. There is a system consisting of a pair of plates oppositely
charged. The charge and geometry define a fixed energy of this system
(assuming it resides in a vacuum). Then a near by electron is
influenced by the system and accelerates between the plates and
emerges outside of the sytem with a velocity much higher than it had
before it entered the system. A snap shot immediately outside the
side of the system, we see the electron with more energy than it had
before it entered the system. The electron has a net increase in
energy. The system (not including the electron) must be at a lower
energy state because the electron has gained some energy. The
geometry and charge on the plates has not changed so it has the same
energy that it did before the electron entered.
So we have the system with no change in energy and the electron has
gained energy. Where did the energy come from? What lost the energy
that the accelerated electron now has?
Ray
John Polasek
wrote:
>On Dec 8, 2:19�pm, John Polasek <jpola...@cfl.rr.com> wrote:
I'm afraid that you follow Smythe and Panofsky withthe same na�ve view
of this capacitor-like arrangement i.e. two plates bearing +Q and -Q,
without asking how they got there. Charge reciting on a grid of wires
is almost infinitesimally small: in Gaussian electrostatics the
capacity is given by the radius of the sphere. The radius of your
wires are microscopic by comparison.
Here is a little write up I did vis-�-vis Panofsky:
Capacitors are represented as two charged metal plates having equal
and opposite charges and surface charge densities +sigma and -sigma.
Faces of inserted dielectric would acquire surface charges of opposite
polarity. A force between the faces is calculated as F/A =
sigma^2/2eps0 (Eq.6-78) "as long as the dielectric doesn't touch the
plates (!)" with no reference to voltage and a clear an extension of
Coulombs equation.
A crucial well-known experiment is illuminating: if 2 plates have
force F and dielectric k is inserted, the force is declared to be
reduced to F/k. Panofsky calls it mysterious and complains "it is
difficult to see on the basis of a field theory why the interaction
between two charges should be dependent on the nature or condition of
the intervening material�1/k lacks a physical explanation". Smythe
also, on pg. 2 defines e0 in Farad/m but fails entirely to acknowledge
any vestige of permittivity (not found in index) or polarizability:
..�
"when sufficiently sensitive methods are used, charge also
exerts a small attractive force on uncharged insulators. This seems to
indicate that even in insulators charges are present and are not
absolutely fixed but may suffer some displacement. The hypotheses
concerning the actual behavior of the electric charges in conductors
and insulators will not be discussed here. The theories are still
imperfect but have greatly improved since 1930�"
The charges got there by connecting a battery (battery is never
mentioned by either of the above authors in their books) to the two
plates with a galvanometer that will show a gulp of charge. Thereafter
the energy resides in the *polarization of the medium* namely the
vacuum that has the property epsilon0 and not at all in the charge on
the plates. You apparently have the view that the electrons are there
like ornaments incapable of being dislodged.
Without going into the details of the equation div D = rho, recall it
is established that when a dielectric is inserted between charged
plate capacitors the terminal voltage drops, or in the case of P&P
that the force between the plates is reduced, (application of the
coulombs law experiment). The reason voltage is reduced is because
with a dielectric (K = 1 + X) present there are X more polarizable
charges then in the vacuum alone. Since E = D/K*epsilon, E and V =
E*g drop.
The inserted dielectric has *bound* charges. By contrast, a conductor
has myriad *free charges* and your electron cloud of unbound charges
becomes a virtual conductor so it should completely or partially short
out the whole thing: the bound charges are able to relax their
polarization without the benefit of an external shorting wire.
I would estimate that if the electrons in your cloud are equal to Q in
the capacitor, it will become totally discharged. Any surplus
electrons would simply become stranded because there would no longer
be any electric field.
Do you have a reference for this experiment? Has it ever been done?
The authors PP&S above have no explanation for reduction of potential
from the dielectric because, while they claimed to welcome the new SI
model, circa 1960, they apparently were dyed in the wool Gaussians who
never really could believe in epsilon and mu of the vacuum.
My reference books Panofsky and Smith date back to the 1960s where
they were the standards. I would be glad to hear if there are newer
books that have a more enlightened view of permittivity.
You might want to look at my permittivity paper on my website where I
have deduced the properties of the so-called quantum vacuum:
Dualspace.net.
John Polasek
Anti Vigilante
>
> That was a fun ride! Thank you.
>
> I would like to address my original posting where I never mentioned
> field.
That's unfortunate because it's a useful tool.
> There is a system consisting of a pair of plates oppositely
> charged.
You have a dam opened on the left side and a waterfall on the right side.
> The charge and geometry define a fixed energy
No. They define a current of virtual photons constantly flowing.
Energy (potential) is a measurement of action that you could later
extract from a particle if you have the opportunity to put it there in
the first place. Potential energy is a measure of what it costs to
prepare to do something. Kinetic energy is a measure of the returned
value as it is extracted.
> Then a near by electron is influenced by the system and accelerates
> between the plates and emerges outside of the sytem with a velocity
> much higher than it had before it entered the system.
So our boat enters the water flow.
> A snap shot immediately outside the side of the
> system, we see the electron with more energy than it had before it
> entered the system. The electron has a net increase in energy. The
> system (not including the electron) must be at a lower energy state
> because the electron has gained some energy.
The waterfall and the dam never lost energy. The water (virtual photons)
in between does.
> The geometry and charge on
> the plates has not changed so it has the same energy that it did before
> the electron entered.
Same non physical potential energy. They have no kinetic energy.
> So we have the system with no change in energy and the electron has
> gained energy. Where did the energy come from? What lost the energy
> that the accelerated electron now has?
The water (virtual photons) gave up their life so you can watch TV. :)
> Ray
Tom Roberts
> I would like to address my original posting where I never mentioned
> field.
You did not mention "field", but the only models that successfully describe this
situation involve a field. Whether you mention it or not.
> There is a system consisting of a pair of plates oppositely
> charged. The charge and geometry define a fixed energy of this system
> (assuming it resides in a vacuum). Then a near by electron is
> influenced by the system and accelerates between the plates and
> emerges outside of the sytem with a velocity much higher than it had
> before it entered the system. A snap shot immediately outside the
> side of the system, we see the electron with more energy than it had
> before it entered the system. The electron has a net increase in
> energy. The system (not including the electron) must be at a lower
> energy state because the electron has gained some energy. The
> geometry and charge on the plates has not changed so it has the same
> energy that it did before the electron entered.
Yes. This shows two aspects of the field due to the charges on the plates AND
THE ELECTRON:
a) the field can accelerate the electron.
b) the field has energy density; the total must decrease as the
electron is accelerated.
> So we have the system with no change in energy and the electron has
> gained energy. Where did the energy come from? What lost the energy
> that the accelerated electron now has?
From the field. It does not matter whether or not YOU mention "field", all of
the models we have for such a situation include an electromagnetic field.
Note that you use different meanings of "system". In classical
electrodynamics, the system consists of the plates, the
charges on the plates, whatever (neutral) supports maintain
the locations of the plates, the electron, the charge on the
electron, and the EM field throughout spacetime. This
system has constant total energy. But when you consider just
a part of the system, that part's energy can vary.
Note that if the plates are not perfectly conducting, the redistribution of
charge in them will generate some heat, which removes energy from the parts of
the system you are considering. As does the radiation generated by the
electron's acceleration and the movement of charges in the plates.
> We need to be able to show how the energy density of the field has
> changed.
This is where solving Maxwell's equations comes in. Consider the instant when
the electron passes through the positive plate (imagine there's a tiny hole just
for this). At that instant, the total EFFECTIVE charge on the positive plate is
reduced by the charge on the electron [#], so the E field nearby is reduced.
That implies lower energy density near the plate, and lower total energy in the
field. Note that the positive charges in the plate will move in response to the
presence of the electron, and the negative charges in the other plate will move
as well. There is also heat generated in the plates, reduced strain in the
supports and plates, and some EM radiation moving away. The reduced total energy
of the field is by far the largest contributor to the change in energy of
everything but the electron.
[#] that is, if you integrate over the volume of the plate,
INCLUDING THE HOLE, the total is reduced. It is this
effective charge that determines the field nearby.
YOU may not care about the field, and may not believe it "exists" (for some
meaning of that word). But somehow you need to describe how the charges on the
plates affect the electron, and how overall energy is conserved. Every model we
have that applies to this situation does have a field, with the properties I
mentioned above.
You are unlikely to succeed in describing and understanding
this without obtaining a good textbook on electrodynamics and
STUDYING it.
Tom Roberts
ray
> ray wrote:
> > I would like to address my original posting where I never mentioned
> > field.
>
> You did not mention "field", but the only models that successfully describe this
> situation involve a field. Whether you mention it or not.
>
> > charged. The charge and geometry define a fixed energy of thissystem
> > (assuming it resides in a vacuum). Then a near by electron is
> > emerges outside of the sytem with a velocity much higher than it had
> > side of thesystem, we see the electron with more energy than it had
> > before it entered thesystem. The electron has a net increase in
> > energy. Thesystem(not including the electron) must be at a lower
> > energy state because the electron has gained some energy. The
> > geometry and charge on the plates has not changed so it has the same
> > energy that it did before the electron entered.
>
> Yes. This shows two aspects of the field due to the charges on the plates AND
> THE ELECTRON:
> a) the field can accelerate the electron.
> b) the field has energy density; the total must decrease as the
> electron is accelerated.
>
> > gained energy. Where did the energy come from? What lost the energy
> > that the accelerated electron now has?
>
> From the field. It does not matter whether or not YOU mention "field", all of
> the models we have for such a situation include an electromagnetic field.
>
> Note that you use different meanings of "system". In classical
> charges on the plates, whatever (neutral) supports maintain
> the locations of the plates, the electron, the charge on the
> electron, and the EM field throughout spacetime. This
> a part of thesystem, that part's energy can vary.
>
> Note that if the plates are not perfectly conducting, the redistribution of
> charge in them will generate some heat, which removes energy from the parts of
> electron'saccelerationand the movement of charges in the plates.
Tom,
The system as you describe it, is the whole universe (all spacetime).
The scientific process requires us to be able to define a subset of
the universe as our system under test.
In this case, the system is a set of plates with a charge and
geometry. That combination defines an amount of energy. Neither the
charge nor the geometry is changed through the experiment which says
that the energy is the same before and after the experiment. We have
an electron which was outside the system under test. The electron had
a set amount of energy at time zero. It then drifts into the space
between the plates where it accelerates and exits. At exit, the
electron has more energy than at time zero. There are many things, as
you pointed out, that go on. But in the end, the system has the same
energy and the environment (at least the electron) has more energy.
Per your description, we have also lost energy thermally in the
plates. But the only source of energy, the charge and geometry, has
not changed - the energy has not changed.
The only way the system can loose energy is to change the geometry
(bring the plates closer together) or decrease the change on the
plates (they are isolate and thus no current flow).
I totally agree that the laws of nature must prevail. I just want to
see how they are expressed.
Ray
Tom Roberts
> The system as you describe it, is the whole universe (all spacetime).
Yes. This is inherent.
> The scientific process requires us to be able to define a subset of
> the universe as our system under test.
Why would you think that??? If you consider only a part of the universe, then in
particular you cannot invoke energy conservation within that part, because it
can exchange energy with the rest of the universe. Ditto for momentum
conservation, and just about any other conservation law of physics.
Your claim is not part of any "scientific process", though it is true that for
many physical situations one can APPROXIMATELY separate out a part of the
universe for analysis, and in many cases the approximation is exceedingly good.
A VERY IMPORTANT aspect of science is to understand what approximations one is
using, and how they might affect one's analysis. You keep trying to use an
approximation (ignoring the EM field) that completely invalidates your analysis.
> In this case, the system is a set of plates with a charge and
> geometry. That combination defines an amount of energy.
No. Just because you don't want to mention "field" does not mean that you can
model this system without an electromagnetic field. Just because you want to
mention only a part of the system and consider it "the system" does not mean
your analysis of that part is complete.
And you forgot the electron -- it is CLEARLY part of the system.
> Neither the
> charge nor the geometry is changed through the experiment which says
> that the energy is the same before and after the experiment.
Only when one considers the energy in the EM field throughout spacetime, plus
the energy of the electron, plus the heat in the plates, plus the strain in the
supports. It is important to enumerate these contributions to the total energy
of the WHOLE system, so one can evaluate whether or not they are important. For
this case, the energy in the electrostatic field is the only important contribution.
> We have
> an electron which was outside the system under test.
The EM field permeates spacetime, and for a complete analysis you cannot put the
electron "outside the system". But you MUST consider the WHOLE system.
> The electron had
> a set amount of energy at time zero. It then drifts into the space
> between the plates where it accelerates and exits. At exit, the
> electron has more energy than at time zero. There are many things, as
> you pointed out, that go on. But in the end, the system has the same
> energy and the environment (at least the electron) has more energy.
But you keep ignoring another aspect of the WHOLE system -- the EM field. That's
why you keep coming up with what appears to you to be a problem. THERE IS NO
PROBLEM, except for your insistence on ignoring part of the WHOLE system.
> The only way the system can loose energy is to change the geometry
> (bring the plates closer together) or decrease the change on the
> plates (they are isolate and thus no current flow).
This is just plain not true. Another way the PART of the system you are
considering can gain or lose energy is by interacting with the EM field. Indeed,
that is precisely how our models of this behave. There are also minor
contributions to the energy that you ignore.
> I totally agree that the laws of nature must prevail. I just want to
> see how they are expressed.
Look in any textbook on electricity and magnetism, or better, on electrodynamics.
This is getting overly repetitive; don't expect me to continue
until you learn something about the subject. Just because you
cannot "see" the field does not mean it isn't needed to model
this system.
Tom Roberts