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Voltage gradient conservation of energy problem
<[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
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
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
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
Linearity for practical devices is largely a
purposely incorporated feature of the chosen
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
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
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
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
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.
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
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.
1. Electrons falling through an electric potential gain
2. The Q in a capacitor is due to charge displacement or
actual charge. Electrons are handy mobile charge
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
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.
|conservation , energy , gradient , problem , voltage|
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