<[Only registered users see links. ]> wrote in message
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Is there a question in the above? There should be, as
the statements are not correct.

Charge is a conserved property, so if the geometrical
configuration and disposition of charges that give
rise to your potential gradient are unchanged, then
the voltage gradient will remain unchanged.

Energy is not destroyed when a charged particle traverses
a voltage gradient; you simply have to re-examine the
extent of what constitutes your closed system. If a
charged particle is accelerated in one direction, the
apparatus that is holding the charges comprising the
gradient will be accelerated in the opposite direction
(and in inverse proportion to the masses involved).

<[Only registered users see links. ]> wrote in message
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A voltage difference is caused by the concentration of
charges. An open circuit does not discharge because
the electrons have no conduction path by which to
escape. This is why, for example, a conductive body
with a static charge holds that charge on its surface --
the electrons move to locations as far from each other
as possible.

It may blow apart if it is not designed for what you are
trying to do. Why do you assume that the 1 farad capacitor
should be correctly designed but not the coil?

And any other ancillary wiring between them.

Anyways, a charge on a capacitor comprises a separation
of charges, with a surplus of electrons on one plate and
a dearth of electrons (leaving a net positive charge due
to the fixed nuclei of the matterial comprising the plate)
on the other. Dialectric effects also come into play as
the material of the dialectric polarizes as well.

Your understanding is faulty. The charge on the central
capacitor is due to charge displacement. Electrons are
"pushed" from the plate of one of the outer capacitors
onto one plate of the central one, and an equal number
of electrons are "pushed" from the opposite plate onto
the plate of the other outer capacitor. The net result
is a voltage difference across the central capacitor.

Electrons do not have to flow through the dialectric of
a capacitor in order for it to work. In fact a good
dielectric should be non conducting but polarizable.
Ultra pure water, which is a strongly polar molecule,
would make an excellent dielectric if it could be kept
from becoming contaminated by ions from the plates and
container.

What do you mean by "voltage energy"? Voltage is a measure
of the potential energy due to an electric field, that is,
the number of Jules of potential energy per coulomb that
a charge located at that given point in the field has.

Also, the electric field is not comprised of electrons
(or protons or any other charged particle). It is the
property of an electric charge that it has a field which
extends throughout space. The carrier of the electric
and magentic field is the photon.

An analogy can be made to a gravitational field. Mass
has a gravitational "charge", and objects in the field
of that charge will have a potential energy due to their
location within that field. We don't expect to see gross
matter flying back and forth to support the potential.

Now you're mixing in the work done by a system, and in
particular, the work done by charges moving in an electric
potential that are affected by friction (resistance), and
the conversion of electrical potential energy to magentic
potential energy in a dynamic system. Why are you
complicating things beyond your original simple thought
experiment with a single electron moving freely in a
static electric field?

What's this got to do with the price of fish in China? How
does it relate to your simple thought experiment and your
original statements?

Voltage is *defined* to be the magnitude of the electric
field (the field potential) which is due to a distribution
of electric charges. The plates are superfluous to the
definition, but it so happens that charged plates will
exhibit an electic field.

Moving charges create a magnetic field. A conductor placed
in a voltage gradient will generate a magnetic field as
long as the current flows. If the conductor is isolated,
the current flow will be brief while the charge carriers in
the conductor sort themselves out, polarizing the conductor
(induced voltage potential across the conductor). After this,
the current dies as does the magnetic field.

<[Only registered users see links. ]> wrote in message
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First, your terminology is incorrect. Voltage represents
an energy potential --- Jules per Coulomb. "Voltage energy"
doesn't really make sense.

The equation is saying that as you reduce the capacitance,
the energy required to maintain the charge separation
rises, and tends to infinity as C --> 0. This makes sense
since it corresponds to squeezing the charges on each plate
into smaller and smaller volume (zero in the limit).

Of course, as the charges are squeezed into ever smaller
volume the field density rises. There will be a spontaneous
breakdown (arc discharge) long before any infinities can
arise.

There's nothing wrong with the equation (barring Relativistic
effects that are not accounted for in extreme circumstances).
Correct interpretation of the equation requires some
understanding of the underlying mechanisms.

Q is defined to be the quantity of charge, and charge is
observed to come in discrete units. Electrons have a
fixed unit charge, as do protons, for example.

Charges are accompanied by an electric field that varies
as the quantity of charge and inversely as the square of
the distance from the charge. Voltage is the measure of
the potential energy that a test charge will have when
placed in an electric field (Joules per Coulomb).

The electric field in a capacitor is the result of
the sum of the fields of the distribution of charges
comprising the capacitor.

Any moving charge will result in a magnetic field.
A circuit is not required.

And heat, and light, and sound energy.

You seem to be fixated on the plates of a capacitor for
some reason. The plates are simply a convenient mechanical
platform for holding the charges and conducting them around
as required. Note that the equation for the capacitor do
not specify anything other than the area of the plates and
their separation. This is only to define the distribution
of the charges, the plates themselves are entirely
superfluous.

If you like, a thundercloud and the ground represent the
plates of a capacitor, too. The cloud is holding charge
on water droplets.

Voltage is the measure of the potential energy that exists
in the system of charged particles. In particular, it is
the specific potential energy of a test charge at a given
location in the field, that is, the number of Joules of
potential energy per unit charge of the test charge.
Voltage isn't a "thing" like charge is; Voltage is a result
of the geometry of the system of charges.

Note that potential can be positive or negative thanks
to the existence of opposite charges.

There are only so many ways that forces can come to bear
on particles. These correspond to the known forces of
nature. The ones that we deal with most in day to day
life are the electromagnetic forces and the gravitational
force. If charges are being held fixed against a potential
then there is some force at play which is doing it. In
most cases this will be found to be other electrostatic
forces, such as the attraction of nearby positively charged
atomic nuclei which are in tern held in place essentially
electrostatically (inter-atomic bonds).

Real world superconductors have current limitations.
They tend to lose their superconducting properties
(often catastrophically) beyond a given current density
or magnetic field intensity.

If you want to say we are considering an ideal superconducting
coil for purposes of discussion then I am okay with that. But
I wish to stipulate that it is made from high grade Unobtainium
and is physically indestructible. Your argument then is moot.

You have it exactly backwards. The capacitor's voltage is
due to the accumulation of charge in the capacitor. The
voltage represents the stored potential energy in the form
of an electric field in the capacitor. The thing that
drives the charges against the growing voltage on the
capacitor is the collapsing magnetic field (magnetically
stored potential energy) of the coil.

Is that a question or a statement?

The potential would have to be pretty high to strip
the conduction band electrons completely away from
the atomic nuclei of a conductor. The energy required
to shift the density of electrons in the conduction
band to create a potential difference is far smaller
than that required to completely remove electrons
from the conduction band, which is what would be
required to see a significant change in the emission
spectra.

No. No closed circuit is required to create a magnetic
field, only moving charge.

A moving electron will create a magnetic field.

No, that's the energy in Joules stored as electrical potential
energy in the capacitor. Another equivalent formulation is:

E = (1/2)*C*V^2

Huh? Why bring this up? I never said anything about
mass energy conversion, and it certainly isn't required
in the present discussion.

Electrons carry electric charge. The electric charge
has an attendant electric field. The field is mediated
by photons. What's the problem?

Voltage is the result of the geometrical disposition of
electric charges. What happens to those charges if there
happens to be conduction paths or not is irrelevant to
the potential that exists at a specific instant.

Superconductors are interesting in that the electrical
resistance is zero. This does not mean that the
inductance or capacitance of the superconductor is also
zero. Even a perfectly straight superconductor one
inch long has inductance.

For a constant current the voltage across a superconductor
is zero. Vary the current and you get the usual inductance
effects and potentials.

I hold with the standard view as per Maxwell's equations
and the accepted definitions of quantities and parameters
as used in electric theory.

If you have some other concept in mind then you'll have to
be very careful that they don't trip you up (lead to
incorrect results) when you apply them to real world
problems.

Well, your belief is at odds with the accepted definition
and usage of the term.

They don't. Electrons are moved physically by the action
of the silk on the rubber. The silk donates electrons to
a given area of rubber, and since it is an insulator, they
tend to stay put where they are placed. The result is a
concentration of negative charge. Any voltage gradient
is strictly relative.

No, it won't, and for the same reason that a bouncing ball
doesn't deplete the gravitational field of the Earth.

If the electron bounces, then it moves back against the
potential gradient and regains some potential energy
(it exchanges kinetic and potential energy). This
potential energy will then be converted back to kinetic
energy when it resumes falling towards the anode.
Rinse, repeat.

Energy will be lost to the system through electromagnetic
radiation as the electron is accelerated, even if the
bouncing is made perfectly elastic.

Electrons are not specifically required, electric charges
are. Protons or ionized atoms or molecules will do
nicely, too.

What is the geometry of the magnetic field produced
by a moving electron? What is the magnitude and
direction of the force due to that field on the moving
charge? The field surrounds the electron and has the
form of concentric loops. The electron passes down
through the center, and the net force is zero. In
other words, the magnetic field due to a moving charge
doesn't self interact with that charge.

More properly, electric and magnetic fields. Electricity
and Magnetism have other connotations.

This is essentially true. The photon-photon interaction
cross section is very small indeed, and effects only become
measurable at very high energy densities indeed.

Electron movement creates a magnetic field. If the
electron is accelerated (wiggled, say) then an
electromagnetic field results.

Photons are the mediator of the electric and magnetic
fields, but are not themselves charged. So they don't
create more fields of their own.

No problem. But you should start thinking in terms of Electric
energy and Magnetic energy rather than voltage energy.

<[Only registered users see links. ]> wrote in message
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Better, but still questionable.

Um, doesn't this contradict your claim above that charge
density doesn't affect voltage?

C ~ A/d and so it's no surprise that the voltage energy
density goes up as the plate separation shrinks. That
means that the field strength (volts per meter) between
the plates must go up.

Charge is the ultimate source of an electric field,
and to achieve a higher field strength you want to
concentrate the charge in as small a volume as
possible so that the r in E = k*Q/r^2 can be
minimized.

The electric potential (in Volts) at a given point in the
field is given by k*Q/r , so the same logic applies.

No one would say that increasing the plate separation causes
an increase in the electrical energy. In fact, the effect
is quite the reverse; it takes mechanical energy to move
similarly charged plates together (they repel). When this is
done, the voltage across the plates increases while the charges
on each stay the same (assuming electrically isolated plates).

Where do you see written above anything about moving
the plates apart?

Arcing is the result of the field strength surpassing the
breakdown strength for a given media. Even vacuum has
such a limit.

The field between the plates of a parallel plate
capacitor is remarkably uniform except for very
close to the edges, provided that the plates
dimensions are relatively large compared to their
separation.

I think you're confused. There's no rule that says that
the potential between charges must be linear with
distance. It just so happens that in many cases of
interest it is so, as with the field between the plates
of a parallel plate capacitor (except near the edges) as
described above.

The potential at a given distance from a spherical charge
is given by:

V = k*Q/r

If we have two spheres with charges Q and -Q and unit
separation (and without loss of generality) then for a
point between their centers (by superposition):

V = k*Q/r - k*Q/(1-r) r is the distance from one
of the centers, so (1-r) is
= k*Q(1/r -1/(1-r)) the distance to the other.

= k*Q(1 - 2r)/(r-r^2)

If you plot this function you'll see that for much
of the region along the path from the center outwards
is quite linear, but grows exponentially close to
each charge.

Linearity for practical devices is largely a
purposely incorporated feature of the chosen
geometry.

What effects? You'd need to provide empirical data on
emission spectra one way or the other.

We can certainly tell how much energy gradient is required
to separate electrons entirely from their parent atoms.
This is given by the energy of the photons in the spectrum,
and is usually in the area of a few electron volts. Chemical
reactions too, for example, deal in energies of this order.

An electric field does not have to completely remove
electrons for there to be a significant induced potential.
It is only neccessary to shift the average position of
the orbitals slightly, exposing a bit more of the nucleus'
field on one side. In effect, the atoms behave like little
electric dipoles, and their effect adds up over the
geometry.

Perhaps a review of Maxwell's equations are in order?

There's a reason it's called "Electromagnetism". Both
electric and magentic fields are in fact part of the
same whole. You might think of a magnetic field as
seeing an electric field from a different "angle",
and vice versa, and that angle is due to the velocity
of the charge. (Note, this is not a real, physical
angle but a conceptual one).

Any velocity other than zero.

The strength of the magnetic field depends upon the velocity.
If E is the electric field of a given charge and v is the
velocity of the charge, then

B = v x (1/c^2)E

is the magnetic field. Note that 'x' represents the
vector cross product, and E is a vector representing
the electric field.

At the atomic level you need to largely abandon classical
electrodynamics and replace them with quantum electrodynamics
(QED). The fact that 'orbiting' electrons didn't lose
energy and cause all atoms to spontaneously collapse was
one of the things that spurred on the investigation into
quantum theory.

It would depend upon whether or not all of the
spins of the electrons and protons were paired.
If they are paired, they all cancel out at a
distance of several atomic radii.

Transistors must be doubly so then.

The energy is stored in the electric field, the electric
field is the result of the geometry of the charge
placement. It just so happens that the plates are
providing a convenient source of charges to push
around.

An electron volt is a unit of energy. It is *equivalent*
to the energy acquired by accelerating an electron
through a potential of one Volt. One volt is equivalent
to one Joule per Coulomb.

Could be. Any distribution of charges will result in
an electric field. The potential at any given location
is the sum of the potentials due to all charges.

There are several failure modes. One of them would be
for the lateral forces caused by a high current density
to exceed the physical capability of the material.
There are other effects depending upon the nature of the
superconductor and the applicable superconduction
model.

You were the one who claimed that a superconductor would
explode if it were forced to carry a high current. Well,
my argument is that the physical limits of any real
component can be exceed. This is an engineering problem,
not a theory problem.

Again, that would depend upon the material and current
density. You could also make a conductor out of a
fluid where the charge carriers are molecular ions
(see electrolysis, for example). Electrons are not the
only charges things lying around.

The magnetic field is due to moving charges, not an
accumulation of charges. The higher the rate of
charge motion (i.e., the higher the current), the
larger the magnetic field.

It would if the intervening space could prevent
breakdown due to the high field stength that would
prevail. Charges would jump the gap.

And Maxwell.

If the elctron is moving linearly, yes. If it is
accelerating in a curve then it will lose energy
in the form of electromagnetic waves (the so-called
synchrotron radiation). This is an important energy
loss mechanism in particle accelerators.

Real photons (electromagnetic waves) are produced when
charges accelerate. In most practical cases the
energy carried off by electromagnetic emission is tiny
compared to the energy in the circuit. You can work
out the energy of the individual photons via the
Planck relation, E = h*f, where f is the frequency.

That's not the Pauli Exclusion Principle, that's the
photoelectric effect and Quantum Electrodynamics.
A Google search on Electromagnetic Field will turn
up plenty of references.

A conduction path merely allows the charges to
move. Moving charges produce the magnetic field.

The magnetic energy is strictly due to the current
and inductance. The induced voltage is proportional
to the rate of change of the current multiplied by
the inductance.

If you apply a voltage potential to a superconductor
(from an ideal voltage source), then the current
will rise linearly and indefinitely, maintaining the
the V = L*(dI/dt) relationship. If this were possible
using real world components, one could store any amount
of magnetic energy in a superconducting coil's field.

Maxwell trumps all other contenders when it comes
to empircal verification under all circumstances.
It's also immediately compatible with Relativity.

Wikipedia is playing a bit fast and loose then. It is in
fact equal to the work required to move an electron through
a potential of one volt. Work can be positive or negative.

If an electron has kinetic energy (which an electron in
motion does thanks to its mass and velocity) then it
can trade this kinetic energy for potential energy in
an electric field. In other words, an electron can
move to a higher potential by expending kinetic energy.

I think:

1. Electrons falling through an electric potential gain
kinetic energy.
2. The Q in a capacitor is due to charge displacement or
actual charge. Electrons are handy mobile charge
carriers.
3. Magnetism is a field resulting from the movement of
charges. That is, it depends upon the current.

Electrons boiled off of the cathode of an electron
tube can certainly travel to the positively charged
plate and reduce the potential there unless it is
maintained by an external source (power supply). If
the electrons travel continuously from cathode to
plate then we say that the tube is conducting and
that there is a plate current (and cathode current).

The isomagnetic lines of force can be mapped around
a current carrying wire, and they appear take the
form of concentric uniform intensities.

It's not that the field strength drops to zero at
the location of the center of the charge, just that
its net affect on the charge sums to zero there.

Not exactly. The electromagnetic field that surrounds an
electric charge (purely electric if the charge is stationary,
a combination of electric and magnetic if in motion) is
mediated by photons. Photons are particles exchanged by
charged particles.

Why should it? Voltage is the just the potential
energy associated with a test particle located in an
electric field. Volt = Joule per Coulomb.

Voltage is an effect of the geometry of the field and
is position dependent. It relates to the amount of
energy that a free falling charge can obtain from that
location in the field. The total energy stored in the
field is another matter.

<[Only registered users see links. ]> wrote in message
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A resistive path is not required. Free space is enough
for an electric field to present a voltage gradient.

It matters to the field immediately around the contacts.
It averages out to a linear progression some distance
from the contact point, and depends upon the specifics
of the geometry of the contact/medium interface.

Look up "edge effect" for parallel plate capacitors.
The smaller the plates in the given geometry (spacing),
the larger the edge effect contribution and the more
non-linear the field between the plates.

Insulators are just relatively high resistances.
Geometry affects the field shape so that voltage
gradients are not always nice linear things.

No one says they do. However, the geometry of the
contacts (or plates) certainly has an effect upon the
shape of the electric field in the device. It is
usually of little or no consequence for your average
electronic part, but becomes critical in high voltage
or high precision circumstances. For example, there's
a reason that the accumulators atop Tesla coils are
spherical rather than pointy.

The above would appear to be information unrelated to
any part of the discussion. Nevertheless, the non-ideal
characteristics of real devices, such as the inductance
and capacitance associated with real-life resistors, are
again a function of the geometry fo the devices.

Sorry about that.

Voltage across a resistor cannot be sustained without
an external supply because the charges are free to
move and cancel. Note, however, that an isolated
resistor can sustain an overall static charge just as
any object can.

Voltage across a capacitor can be sustained without
an external source because it stores energy. If you
charge a capacitor and then remove the supply, it
retains the applied voltage. If you then vary the
plate separation or the plate size, the voltage will
vary according the the relevant equation.

Classically, the field does become infinite near a
point charge. In the real world quantum effects
rule at small distances.

E = q/(4*pi*e0) * r/|r|^3

where r is the distance vector from the charge q.

This would only be true if the plates were infinite
in extent. For a real capacitor the plates are of
finite dimension. As they separate, the field between
the plates becomes increasingly non-uniform (the
edge effects again), and as the separation distance
exceeds the plate dimensions the plates begin to look
more and more like point charges. At large separations
the potential energy will be indistinguishible from
that of two opposite charges separated by the same
distance.

Electrical and mechanical energy are interchangeable.
Weren't you the one considering an electron falling
in an electric field?

You mean:

I took that statement to indicate that the plates would be
removed, somehow leaving the naked charges in place.

Besides, I was obviously (I thought) referring to the
paragraph above your query about whether moving the plates
would cause arcing.

Breakdowns occur at locations of highest field strength.
As you surmise, this is closely associated with the geometry
of the material; smooth good, pointy bad.

It does and it is. Look it up; there are many publications
that discuss vacuum breakdown, particularly as it applies
to vacuum insulated spark gaps.

Charge on an isolated conductor tends to lie on its
surface, so the thickness is irrelevant in that case;
the charge will be equally distributed on the face of
the isolated cube.

For a charged capacitor situation, the opposite charges
attract and charge will migrate towards the facing
sides, but will remain on the surface (even the free
charges not on the facing sides).

You should keep in mind that the cubes of material
themselves will have what is known as self-capacitance,
too, which will complicate the calculation of the
overall capacitance of such a capacitor. Again, it
comes down to geometry.

If you increase the charge density at the ends of the
wire then the field density there will be greater.
Increasing the charge density means shrinking the
contact area. Resistance, too, depends upon geometry,
so that if the contact area shrinks the conductivity
in the vicinity of the contacts shrinks too.

So what, in your view, is the mechanism of a battery
if it is not charge separation by chemical process?

That's the classical result, yes. You may be hard
pressed to find a real sphere with radius zero though,
and quantum theory steps in as you approach the size
of an atom.

Do batteries in series add?

Now you're changing your tune and saying that Farads
are not based on electrons. Before it was charge.

Capacitance can be looked at as being a proportionality
constant relating charge and voltage potential for a
given geometry. V = Q/C. If you muck about with the
geometry then the capacitance changes, which shouldn't
be surprising. It should also not be surprising that
simplifying assumptions used for certain situations
(such as that of a perfectly uniform field between
the plates of a parallel capacitor) break down if the
situation is changed.

References?

I don't understand what you're going on about (or
rather why you're going on about it!). Obviously
the energy of a magnetic field in space is not
zero in that volume of space. Who said it was?
References?

And what's your point about the current in a transformer?
It's a simple application of circuit theory, which
in turn are based upon Maxwell's laws.

The vacuum is polarizable. That's why a characteristic
property of space is e0.

Too bad. Emprically, charges in motion cause magnetic
fields. If you argue with reality, reality always wins.

Photons mediate the electromagnetic field, they
don't initiate it.

Nonsense. Look up the drift velocity of electrons
in a current carrying wire. Quantity makes up for
velocity.

You are wrong. Where's the magnetic monopole in v or E?
B is the magnetic field, not a magnet!

Insulating layers can fail under pressure.

Current density should not be confused with charge
density. The overall net charge on a conductor can
remain near zero while the charge density is very
high.

Superconductors can revert to nonsuperconductors in
the presence of too-high magnetic fields (too
high current density).

Energy is stored in a coil as a result of current.
It is highest when current is highest, zero when
current is zero. Another way to put it is that
the current in the coil is maximum when the voltage
on the capacitor is zero (or passing through zero).

A charge of a Coulomb is a lot of charge, granted,
but you're making assumptions about the physical
properties of components for no good reason.
You can make a coil as large as you want so that
the stresses will be kept within desired parameters.
It's an engineering problem, not a physics problem.

Again, you're pointing at specific choices of circumstances
and components and claiming a that there's a physics problem
rather than an engineering problem.

If a capacitor of a certain physical size can be constructed
to hold a given charge, so can a coil be built to handle the
situation. Since it's current that matters for the coil,
the charge can be spread out over the entire length of the
conductor (as well as any conductor comprising the rest
of the circuit up to and including the capcitor).

<[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

A resistive path is not required. Free space is enough
for an electric field to present a voltage gradient.

It matters to the field immediately around the contacts.
It averages out to a linear progression some distance
from the contact point, and depends upon the specifics
of the geometry of the contact/medium interface.

Look up "edge effect" for parallel plate capacitors.
The smaller the plates in the given geometry (spacing),
the larger the edge effect contribution and the more
non-linear the field between the plates.

Insulators are just relatively high resistances.
Geometry affects the field shape so that voltage
gradients are not always nice linear things.

No one says they do. However, the geometry of the
contacts (or plates) certainly has an effect upon the
shape of the electric field in the device. It is
usually of little or no consequence for your average
electronic part, but becomes critical in high voltage
or high precision circumstances. For example, there's
a reason that the accumulators atop Tesla coils are
spherical rather than pointy.

The above would appear to be information unrelated to
any part of the discussion. Nevertheless, the non-ideal
characteristics of real devices, such as the inductance
and capacitance associated with real-life resistors, are
again a function of the geometry fo the devices.

Sorry about that.

Voltage across a resistor cannot be sustained without
an external supply because the charges are free to
move and cancel. Note, however, that an isolated
resistor can sustain an overall static charge just as
any object can.

Voltage across a capacitor can be sustained without
an external source because it stores energy. If you
charge a capacitor and then remove the supply, it
retains the applied voltage. If you then vary the
plate separation or the plate size, the voltage will
vary according the the relevant equation.

Classically, the field does become infinite near a
point charge. In the real world quantum effects
rule at small distances.

E = q/(4*pi*e0) * r/|r|^3

where r is the distance vector from the charge q.

This would only be true if the plates were infinite
in extent. For a real capacitor the plates are of
finite dimension. As they separate, the field between
the plates becomes increasingly non-uniform (the
edge effects again), and as the separation distance
exceeds the plate dimensions the plates begin to look
more and more like point charges. At large separations
the potential energy will be indistinguishible from
that of two opposite charges separated by the same
distance.

Electrical and mechanical energy are interchangeable.
Weren't you the one considering an electron falling
in an electric field?

You mean:

I took that statement to indicate that the plates would be
removed, somehow leaving the naked charges in place.

Besides, I was obviously (I thought) referring to the
paragraph above your query about whether moving the plates
would cause arcing.

Breakdowns occur at locations of highest field strength.
As you surmise, this is closely associated with the geometry
of the material; smooth good, pointy bad.

It does and it is. Look it up; there are many publications
that discuss vacuum breakdown, particularly as it applies
to vacuum insulated spark gaps.

Charge on an isolated conductor tends to lie on its
surface, so the thickness is irrelevant in that case;
the charge will be equally distributed on the face of
the isolated cube.

For a charged capacitor situation, the opposite charges
attract and charge will migrate towards the facing
sides, but will remain on the surface (even the free
charges not on the facing sides).

You should keep in mind that the cubes of material
themselves will have what is known as self-capacitance,
too, which will complicate the calculation of the
overall capacitance of such a capacitor. Again, it
comes down to geometry.

If you increase the charge density at the ends of the
wire then the field density there will be greater.
Increasing the charge density means shrinking the
contact area. Resistance, too, depends upon geometry,
so that if the contact area shrinks the conductivity
in the vicinity of the contacts shrinks too.

So what, in your view, is the mechanism of a battery
if it is not charge separation by chemical process?

That's the classical result, yes. You may be hard
pressed to find a real sphere with radius zero though,
and quantum theory steps in as you approach the size
of an atom.

Do batteries in series add?

Now you're changing your tune and saying that Farads
are not based on electrons. Before it was charge.

Capacitance can be looked at as being a proportionality
constant relating charge and voltage potential for a
given geometry. V = Q/C. If you muck about with the
geometry then the capacitance changes, which shouldn't
be surprising. It should also not be surprising that
simplifying assumptions used for certain situations
(such as that of a perfectly uniform field between
the plates of a parallel capacitor) break down if the
situation is changed.

References?

I don't understand what you're going on about (or
rather why you're going on about it!). Obviously
the energy of a magnetic field in space is not
zero in that volume of space. Who said it was?
References?

And what's your point about the current in a transformer?
It's a simple application of circuit theory, which
in turn are based upon Maxwell's laws.

The vacuum is polarizable. That's why a characteristic
property of space is e0.

Too bad. Emprically, charges in motion cause magnetic
fields. If you argue with reality, reality always wins.

Photons mediate the electromagnetic field, they
don't initiate it.

Nonsense. Look up the drift velocity of electrons
in a current carrying wire. Quantity makes up for
velocity.

You are wrong. Where's the magnetic monopole in v or E?
B is the magnetic field, not a magnet!

Insulating layers can fail under pressure.

Current density should not be confused with charge
density. The overall net charge on a conductor can
remain near zero while the charge density is very
high.

Superconductors can revert to nonsuperconductors in
the presence of too-high magnetic fields (too
high current density).

Energy is stored in a coil as a result of current.
It is highest when current is highest, zero when
current is zero. Another way to put it is that
the current in the coil is maximum when the voltage
on the capacitor is zero (or passing through zero).

A charge of a Coulomb is a lot of charge, granted,
but you're making assumptions about the physical
properties of components for no good reason.
You can make a coil as large as you want so that
the stresses will be kept within desired parameters.
It's an engineering problem, not a physics problem.

Again, you're pointing at specific choices of circumstances
and components and claiming a that there's a physics problem
rather than an engineering problem.

If a capacitor of a certain physical size can be constructed
to hold a given charge, so can a coil be built to handle the
situation. Since it's current that matters for the coil,
the charge can be spread out over the entire length of the
conductor (as well as any conductor comprising the rest
of the circuit up to and including the capcitor).