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Electric field distribution in a conductor with moving charges

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LAN_403

parsec

Tue Aug 04 2009, 07:17AM

Registered Member #2266 Joined: Tue Aug 04 2009, 07:09AM
Location:
Posts: 3

Hi,

I'm glad i've stumbled on this forum. I've got an unresolved question I've been trying to come to grips with for a while;

I'm trying to reconcile some concepts in electrostatics and dynamics to do with electric fields in conductors with moving charges.

The relative permittivity (constant) of a material seems to be determined by it's ability to be polarized and subsequently induce an electric field within a conductor or dielectric. The permittivity (constant) of a vacuum seems to merely facilitate the calculation of the electric field (voltage) at some distance from a charge or group of charges.

When looking at the electric field in a vacuum some distance from point charge/s, there doesn't seem to be any sort of attenuative effect other than the electric field strength diminshing due to the distance from the point charge (characterised by a the ubiquitous steradian 1/(4PIr^2) function).

I guess the implication here is that lower permittivity constants in dielectrics and insulators result in the speed of electromagnetic wave propagation through these media being lower than the speed of light.

The implication here is that the electric field can propagate through any medium in the absence of motion of charge carriers. This seems counterintuitive though.

Imagine for instance a voltage source (say a bunch of static charges) connected to a strong dielectric (so that the charge cannot spread along the wire) and a long wire. Can you measure the voltage at the end of this wire using a device with an infinite input impedance? I would have thought that you need a small amount of charge to flow into the measuring device to be able to ascertain the electric field at the end of the wire.

Conventional circuit theory would tell you that the voltage at the end of the wire is the same as the voltage at the start of the wire in the absence of moving charges. However, if you consider the attenuation of the electic field due to permittivity considerations, one would expect the voltage at the end of the wire to be lower as the electric field is attenuated by the conductor.

I'm struggling to find an intuitive explantion of how potential distributes itself in media other than a vacuum. What does it mean to say that the potential drops across a circuit loop proportionally based on the current flowing and resistance of the loop?

How are charge carriers imbued with potential that they can shed through collisional processes whilst drifting through a conductor? Assuming that the charge carriers themselves have potential that can be lost seems incongruent with the conductor having an electric field distribution through it based on it's permittivity.

The concept is made even more confusing when you consider an AC circuit as a transmission line (where the wave speed through the circuit is comparable to the alternating frequency, thus creating a voltage gradient purely due to the wave voltage being different at the start and end of the line). It's even more confusing with a reactive load causing the voltage and current to be out of phase.

Help! thanks

Sulaiman

Tue Aug 04 2009, 05:54PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Personally, unless I consider vacuum as non-nothingness (aether etc.)
then it's impossible to form a conceptual model of electromagnetism.
I can't answer all of your questions, but here are a few of my interpretations;
_________________________________________________ ______________
Imagine for instance a voltage source (say a bunch of static charges) connected to a strong dielectric (so that the charge cannot spread along the wire) and a long wire. Can you measure the voltage at the end of this wire using a device with an infinite input impedance? I would have thought that you need a small amount of charge to flow into the measuring device to be able to ascertain the electric field at the end of the wire.

Conventional circuit theory would tell you that the voltage at the end of the wire is the same as the voltage at the start of the wire in the absence of moving charges. However, if you consider the attenuation of the electic field due to permittivity considerations, one would expect the voltage at the end of the wire to be lower as the electric field is attenuated by the conductor.

I'm struggling to find an intuitive explantion of how potential distributes itself in media other than a vacuum.
.......................................... ............................
Imagine your 'strong dielectric' initially with no charge on the surface,
between this surface and the wire is capacitance, I'll call it C1
and between the wire and the 'earth' is capacitance C2.
When the charge is applied to the dielectric surface
an EFFECTIVE current will flow from the surface through the dielectric through the wire and surrounding air/vacuum/dielectric to 'earth'
this is 'displacement' current and it represents the polarisation of dielectrics,
it is NOT a flow of charge-carriers. (except in the wire)
(the internal permittivity of copper is 1, but since it's a conductor it's irrelevant here)
The voltage on the wire (if no current is drawn) will be the voltage on the dielectric
x C1 / (C1 + C2) which may not be entirely intuitive!
One way to almost 'visualise' this is,
accept that for a capacitor dQ = C x dV,
change in charge = capacitance x change in voltage
The 'displacement' current will 'flow' for a finite time then stop
so a finite charge (dQ) 'flows' and
dV1 = dQ/C1 where V1 is the voltage across C1
dV2 = dQ/C2 where V2 is the voltage across C2
dV3 = dV1 + dV2 where V3 is the voltage from the charged dielectric to 'earth'
so
dV3 = dQ/C1 + dQ/C2 = dQ[C1 + C2)/(C1 x C2)]
and
dV2/dV3 = {(dQ/C2) / [dQ(C1 + C2)/(C1 x C2)] } = C1 / (C1 + C2)
hope that wasn't too bad.

As far as I am aware, to measure voltage (ac or dc/static) a current needs to flow
even if only a miniscule 'displacement' current. (pockels cell, field mill etc.)

________________________________________________ ____________
How are charge carriers imbued with potential that they can shed through collisional processes whilst drifting through a conductor?
................................................. ...................
Considering the mobile electrons as ball-bearings
and the wire as a long channel like this |___________________|
e.g. a frictionless plank with side rails
and the atoms of the wire as nails hammered half way in
and perfectly elastic collisions between the balls and nails,
the height of each end of the plank represents it's potential.
At each end imagine an infinite 'ball reservoir'
Whatever balls are on the plank will stay still.
If the potential at one end of the plank is raised above the potential at the other end, each ball will roll 'downhill' as soon as the bit of plank below it is inclined.
The balls will bounce off the nails as they travel, the speed of each ball will reach a kind of 'terminal velocity' which is like the drift velocity of electrons in a conductor.
Tilt the plank more (increase the potential difference) and the overall drift velocity will increase accordingly. ........ I = V/R !!!!!!
_________________________________________________ ______________
The concept is made even more confusing when you consider an AC circuit as a transmission line (where the wave speed through the circuit is comparable to the alternating frequency, thus creating a voltage gradient purely due to the wave voltage being different at the start and end of the line) .
................................................ ..............
Now imagine that the plank is flexible and very long
If one end of the plank is alternately rasied above then lowered below the other end so that a sort of wave travels along the plank,
imagine how the balls will behave ......

Time for tea!

parsec

Thu Aug 06 2009, 01:00PM

Registered Member #2266 Joined: Tue Aug 04 2009, 07:09AM
Location:
Posts: 3

So does a voltage wave propagate through a conductor by polarising charges or like a free electromagnetic wave (at the speed of light, much like it does in a vacuum)?

If it does indeed propagate as a wave (irrespective of the charges in the conductor) then what is it about current flowing through a resistance that causes the voltage to drop?

Sorry, I know this might be a bit ambiguous, I'll try to read up more and perhaps draw a diagram.

thanks

parsec

Thu Aug 06 2009, 01:18PM

Registered Member #2266 Joined: Tue Aug 04 2009, 07:09AM
Location:
Posts: 3

"Considering the mobile electrons as ball-bearings
and the wire as a long channel like this |___________________|
e.g. a frictionless plank with side rails
and the atoms of the wire as nails hammered half way in
and perfectly elastic collisions between the balls and nails,
the height of each end of the plank represents it's potential.
At each end imagine an infinite 'ball reservoir'
Whatever balls are on the plank will stay still.
If the potential at one end of the plank is raised above the potential at the other end, each ball will roll 'downhill' as soon as the bit of plank below it is inclined.
The balls will bounce off the nails as they travel, the speed of each ball will reach a kind of 'terminal velocity' which is like the drift velocity of electrons in a conductor.
Tilt the plank more (increase the potential difference) and the overall drift velocity will increase accordingly. ........ I = V/R !!!!!!"

What do the balls lose when they collide with the atoms though? It can't be kinetic energy as that would change the drift velocity of the charges.

Sulaiman

Thu Aug 06 2009, 05:27PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

The electromagnetic 'wave' moves through the copper exactly as if in a vacuum,
the relative dielectric constant of copper is 1 (I think)
In my model (which I only made up when I read your question so it's a weak model)
the electromagnetic wave propagation is equivalent to the movement of the plank,
not the balls (charges).

In my model if you add more nails (resistance)
for a given potential difference between the ends
the drift velocity of the balls will be lower
so less balls (charge) per second will flow (current).

I don't know how far this model of mine represents reality because
current (!) physics doesn't have any physical medium (the plank) present in a vacuum, this is why I consider vacuum to be something (let's call it aether).
There is no plank (spoon

)
My concept of the universe is that everywhere is a dense medium (aether)
and all of what we call reality is just tiny perturbations about the mean.
I'll stop here before everyone realizes that I'm mad.

I'm just an engineer,
I think we need a quantum physicist to give the 'real' answer.

WaveRider

Thu Aug 06 2009, 07:35PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

First, let is clarify: there is no such thing as a "voltage wave" in a conductor. When electric fields change with time, these need to be represented by a four-dimensional potential (scalar and vector potentials).

An electromagnetic wave cannot propagate easily through a metal because there are "free charges (electrons)" that redistribute themselves in the presence of an applied electric field such that there is minimal penetration of the field into the conductor. For good conductors like copper, silver, etc. this mechanism applies all the way into the visible light spectrum.

The "nails and balls" model is a pretty good one for DC behaviour, but it does not work too well for the AC case.

A metallic conductor is an example of a "degenerate electron gas" and has some similarities to neutral plasmas. Electrons are free to move and the positively charged metal ion are fixed in the structure of the metal. The electrons collide with these fixed charge centers, thereby giving up some of their kinetic energy drawn from the applied potential difference, giving rise to resistance.

BTW Sulaiman- Engineers have won Nobel Prizes in physics as well!

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