sim needs faraday cage?

[happyjack27]

alright, so i have a neutral inner image coil, who's radius and amp-turns i still have to calculate as a function of the magrid coils.

and an outer mirror coil with, conversely, an e field but no m field, to simulate a faraday cage at 3r.

lets say r=0.5 meters center-to-midplane on a wb-6 configuration.
i can calculate the outer mirrors coil charge from erblo's formula. that leaves:

magrid linear charge density <<This should not be linear. It should vary to make a uniform voltage. Maybe something more like [(1+cos(2θ))/2] so that the charge density would be 1/2 where the coils are closest, and 1 at the 45degree point.>>
magrid amp-turns
...

well i have line segments. and for each individual segment i can only set a constant linear charge density.. i could approximate that w/the line segments by adjusting the charge scale factor on each segment.

[happyjack27]

Please note he said "outside the electrons position" which I take to mean effectively outside the ground state. You all are using the Faraday cage to create a ground state. Placing one outside yours will have no effect.

sorry i thought u meant inside the cage. yeah, outside the voltage is constantly 0, of course, since the cage is 0 and there are no charge sources outside it. i intend to model any particles that leave the "confines" of the virtual cage as a loss.

[happyjack27]

on the charge / voltage distribution thing on a conducting surface: bear in mind that charge and voltage are not the same thing, mathematically. the charges on the surface of the conductor will move to make the voltage even across the surface. if you bring a charge source close to the conductor, that means charge sources in the conductor must move towards or away from that source to keep the voltage evenly distributed on the surface. charges always move to even out voltage distributions, that's why electrical circuits work.

[kcdodd]

Can you not just make a segmented box as well? I bet if you made the box big enough you could even use a solution for a point charge at the center of the coils to find an approximate charge density on the box analytically in terms of the grid charge, and then solve for the grid voltage. Or if you want just a further approximation use a sphere to surround the coils so the distribution is uniform. Please, no jokes about how physicists turn everything into a sphere. And then the effect will simply to make the grids have higher capacitance. Where the new condition is that the total charge on the grid should be approximately: Q = V*(1/R_grid - 1/R_chamber)^{-1}/k.

[happyjack27]

hasn't someone already done the analytics? i recall seeing something on ephi. searching through the archives i only found this: (scroll to the bottom)

http://www.mare.ee/indrek/ephi/pef/

doesn't look to be a lot of radial charge redistribution.

i would think it's, in any case, not as uneven as half where the coils come closest, because even where they are farthest apart, they're still fairly close (as opposed to infinitely far, as the 1/2:1 ratio would suggest - the 1/2 to 1 suggests infinitely close:infinitely far) so i wouldn't expect the charge distribution variation to be nearly as pronounced.

in any case, scaling the segments is a lot of meticulous work, so i'm not going to bother with it unless i can get something more precise. preferably something analytically validated. but from that link, at least, it doesn't look like it'd make a very big difference.

[kcdodd]

I'm not sure I understand what you mean by scaling segments.

[happyjack27]

oh, each line segment is defined by a set of numbers, namely:

{x_start, y_start, z_start, x_end, y_end, z_end, current_multiplier, charge_multiplier}

sliders in the upper left of the sim control the "global" current multiplier and the "global" charge multiplier. the linear charge density of a given line segment is simply: global charge multiplier * (that segment's) charge multiplier. so that last number in the data line, "charge_multiplier", simply scales the linear charge density of that wire relative to that of other wires.

so by "scaling" i mean an arbitrary multiplying factor on the charge density, relative to a baseline value that can be changed while the sim is running.

for example, to have a pair of parrallel wires of opposite charge, and no current, one would simply write:

-1, 0, 0, 1, 0, 0, 0, 1
-1, 1, 0, 1, 1, 0, 0,-1

and if you want the charge on the second wire twice as strong as the first, you write:

-1, 0, 0, 1, 0, 0, 0, 1
-1, 1, 0, 1, 1, 0, 0,-2

in truth there are more scaling constants in the segment file. a set to scale all wires by, and then you can create groups of wires and scale the charge, current, and spatial scale on all the wires in a group at once. that's how i'm going to do the image coils. copy and paste, change 3 numbers.

it's nice to be both the end user and the programmer. the programmer always writes for me any nifty little feature i want. and the end user is always so competent in the use of my programs.

[D Tibbets]

I think the confusion is at least in part from using a 'ground state' for the Faraday cage. A Faraday cage certainly needs to be grounded to some huge sink that will not change it's potential in any measurable way despite what current you pour into it. But ground- the Earth is at ground compared to what? It is floating, and is fairly well insulated (the solar wind connects to other areas of the local universe but this is a stupendously high Ohmic resistance compared to the charge stored in the Earth, so it can be considered as an effective ground state, but remember this is only a relative ground.
Thus a conducting plate connected to this 'ground' will not develop a induced bipolar state irregardless of the number and position of point charges on either side of it. The plate has to be insulated from ground for this to occur.

Thinking of a Faraday cage as a ground state is misleading. You could assign any voltage to the 'ground state', positive and negative charges only have significance to this 'grounded reference. This is used in the Polywell when a virtual anode is mentioned. The voltage is still negative compared to absolute 'ground state' of the lab, but it is relatively more positive than the most negative portion of the potential well.

A Faraday cage has absolutely no electrostatic effect on charged particles within it (due to Gauss Law). It is not a ground- that would attract or repel the charged particle depending on the relative voltages. There is no true zero point, only relative points. The charged particles inside the hollow conductive sphere/ Faraday cage are completely independent from the cage and everything outside it * This is because all of the point charges you can designate on the surface of the sphere pull or push on the internal point charge, but these surface charges surround the internal point charge and they cancel each other out. The net effect is zero and thus effectively nonexistent. This is easiest to demonstrate with a closed sphere, but any shape will do,. Even if portions of the wall are not in direct line of sight of the internal point charge, the effect remains the same. At least this is the appreciation I gained from the mentioned previous thread.

Consider transformers. You can stack them and the ground to Earth is the baseline, but so long as you let the ' ground float' each succeeding transformer can multiply the final voltage but only experience the potential exposure between it's output and it's floating (or relative) ground. Assuming there is no insulation breakdown, ideally (not practically) you could stack 1000 V transformers to achieve a final voltage of 100,000,000,000,000,000 volts above Earth ground. Each transformer is acting within it's own universe. The effective ground is not at a static ground, but the 'ground' that floats between the individual pairs of transformers.

To repeat, the Faraday cage does not serve as a ground to the internal point charges but as the border of their universe. Of course once the point charge touches the sphere surface things change. And outside of the cage (Faraday = Magrid) the rules change . The charges are all on one side of the free point charge, but even then there is a transition before the normal inverse square law manifests, again due to Gauss Law. This is explained in the lecture when he talks about a infinite conductive plane that is grounded (Earth ground) or held at some constant voltage with a power supply.

* Gauss laws and the approach to an ideal isolation is compromised when you start to add holes, but it still can be a vastly dominate to mildly significant effect based on the relative hole sizes compared to the total surface area, and other considerations like dV/ dt (voltage rise times divided by time interval ( ie frequency or pulse duration).

I should mention that from an external viewpoint the Faraday cage does act as a ground (to Earth). It will intercept any charged particle or propagating electrical field and absorb it provided the ground is robust enough to to quickly transfer the charge/ energy to ground in a timely manner.
From the outside the Faraday cage is a protective ground, but for an internal free point charge it is the border of the universe. The cage and anything outside of it is nonexistent from an electrostatic viewpoint.

Duh... Let me modify the previous paragraph. The Faraday cage (magrid in this case) is a ground from either direction once a charged particle or electrical filed impinges upon it. But so long as the internal particles are not touching the Faraday cage, it isolates them from any electrostatic influence from outside(and the cage is considered to be on the outside). It would do so whether it was held at Earth ground or any arbitrary voltage (like +12,000 V relative to Earth ground as in WB6).

Dan Tibbets

[kcdodd]

So the simulation doesn't actually solve for the real charge densities?

[kcdodd]

A Faraday cage has absolutely no electrostatic effect on charged particles within it (due to Gauss Law). It is not a ground- that would attract or repel the charged particle depending on the relative voltages. There is no true zero point, only relative points. The charged particles inside the hollow conductive sphere/ Faraday cage are completely independent from the cage and everything outside it * This is because all of the point charges you can designate on the surface of the sphere pull or push on the internal point charge, but these surface charges surround the internal point charge and they cancel each other out. The net effect is zero and thus effectively nonexistent. This is easiest to demonstrate with a closed sphere, but any shape will do,. Even if portions of the wall are not in direct line of sight of the internal point charge, the effect remains the same. At least this is the appreciation I gained from the mentioned previous thread.

This simply is not true. I will try to create a simple thought experiment to show you. Imagine first a spherical shell with a uniform charge density on its surface, and a single point charge somewhere inside the shell. From the shell theorem we know the electric field due to the shell is zero inside. Thus the voltage at the surface of the shell is the superposition of potential due to the shell and of the point charge. One could easily make an analytically formula for this with the origin at the center of the shell (Q) and point particle (q) displaced along the z-axis: V = kQ/R + kq/sqrt[R^2+z^2-2*R*z*cos(theta)]. And theta is the angle away from the z-axis. Thus one can see the potential depends on theta and is not constant on the shell's surface.

Now replace the shell with a conducting shell. This mandates that the potential be the same, and so independent of theta. The point particle charge did not change, thus the charge distribution on the shell must change. If it was spherically symmetric before it cannot be afterward. Therefore the shell theorem no longer applies as spherical symmetry of the charge distribution is broken.

One can qualitatively state that the induced charge distribution creates an electric field inside the sphere which then interacts with the point charge. If the point charge moves up the z-axis, the opposite charge must be induced in the positive z hemisphere, which will pull the point charge further in that direction.

[happyjack27]

So the simulation doesn't actually solve for the real charge densities?

not mine, no. you pre-specify the linear charge density of segments. they're not real conductors, just charge and current sources.

it's designed to be extremely fast, so that i can get a large number of particles and segments in. i figure all that stuff can be pre-calculated if need be.

[happyjack27]

To repeat, the Faraday cage does not serve as a ground to the internal point charges but as the border of their universe.

and does the voltage not drop to zero near the border of the universe? it's not really confined (quantumly) if it doesn't.

[happyjack27]

i beleive what me and kcdodd are trying to state is that a point charge approaching a conductor will act different than a point charge approaching a vaccum. it's like i you have two electrons in free space, or one. the two electrons are going to interact w/each other by following the voltage gradients they produce for each other. a single particle, on the other hand, has no voltage gradient to follow. a conductor has charged particles in it, that can move about freely w/in the conductor. and those particles are going to follow any voltage gradient they see.

yes, their is no voltage gradient inside a HOLLOW sphere. but this is NOT a hollow sphere. there's something in it. there's a charged ball in the center of this sphere. if the ball had the right charge density, you could recover the hollow sphere case. a floating conductor would do for that. but that is not the case here. ever see one of those static electricity balls that arc inside? a sphere inside a sphere. the only reason the inside arcs to the outside is cause it has a different electric potential. if you reverse the case, you can then reverse the direction of the arc. take all the air out and now instead of arcing, you have plasma flow. in either case you're creating an electric current. and that implies a voltage gradient.

a hollow sphere works the way it does because of the r^2 decay of electric potential. if you put a charged ball in that, the ball is _blocking_ that r^2 decay.

try putting a ball cathode inside a sphere anode (use a wire and a small hole). fill it w/salt water or something so the air conducts. see if it produces a current. i bet it does. if it produces a current, that means there's a voltage gradient. now vary the charges on the anode and/or cathode. does that change the current? yes. rule of electric circuits: I=RV. that implies a change in the voltage gradient.

right now if i put a floating faraday cage at V=1, it will acquire a charge so as to not change the voltage gradient. as soon as i ground it, i am changing the voltage gradient inside the non-hollow sphere. i bet you if you had a positively charge free moving ball in there it would suddenly snap to the faraday cage, because by grounding it i'm essentially applying a strong negative field to it. and there's a strong positive field in the center pushing it away, and no longer a positive field on the outside balancing that force.

and how the hell does a fusor work if a charge differential between an inner and outer sphere don't create a voltage gradient?

[D Tibbets]

You should argue with the expert lecturer I recommended, or at least watch the video. Despite this, I'll continue to argue. What you are saying is essentially that Gauss Law is invalid. That will disappoint a lot of people, and require rewriting of Maxwell's equations as Gauss law is a part/ restating of a portion of Maxwell' equations.

I never said a point charge (or smaller sphere charge for that matter) inside a spherical conducting sphere would not effect the larger 'external' sphere. If off center this free point charge would cause induction in the outer sphere surface which would be dampened by a good ground/ regulating power supply but not totally eliminated (even good conduction is not instantaneous, though the e- field changes are (at the speed of light)).
What is not affected is the point charge within the reference sphere. The motion of the point charge would only be effected by other fixed or floating point charges within the surrounding sphere. When the internal point charge moves towards one side, it will induce an opposite charge on the inside surface of the surrounding sphere, and thus attract the floating point charge. But this is where Gauss Law comes into play. As the point charge moves towards one side of the wall, it is pulling away from the opposite side of the spherical wall. This also results in an induced change change but it is weaker than the closer wall. But there is a proportionate greater surface area behind the particle, and this surface area behind the particle which has a weaker induced relative charge, makes up for it by the proportionate greater area of tugging. The opposing forces cancel out. The proportionate increasing area with decreasing interacting field strengths per unit of area perfectly matches the inverse square law changes as the point charge approaches one side of the spherical wall.

A comparison that was made in the lecture to gravitation. Inside a hollow planet there is no gravity. Even outside the sphere consideration of the distribution of attractive masses or forces has consequences for figuring surface gravity or acceleration in both gravity or electrostatics. The Moon's mass is only ~ 1 % of the Earth's mass, yet the surface gravity on the Moon is ~ 17% that of Earth. The discrepancy is because of the radius of the two bodies. The radius of curvature of the Moon is less than the Earth's, thus more of the mass is pulling you down. On Earth, more of the mass is to the sides of you, so it tugs you both ways and cancels out. On smaller bodies like asteroids, the mass may be only 1 billionth of Earth's but the surface gravity may be multiples of this because most of the mass is beneath your feet. In the lecture the talk of electrical fields and acceleration with an infinite charged plane is comparable. I used to wonder why this seemed different from gravity which seems to adhere to the inverse square law much better. But it occurs to me that the scale of the effects is because of the vast difference in the strength of the gravitation force versus the electrical force. A 2 meter square of a charged plate effect on a free point charge a few centimeters from the center of the plates surface is approaching the conditions that would apply if the plate was actually infinitely large. With gravity the plate might need to be several billion meters wide before similar behavior was seen by a nearby particle. I include this information to stress that the strength of an attractive (or repulsive) force is dependent on distance, but also on location relative to the test particle in combination with all other fixed points of the system.

Free mobile particles are a different situation, but even they can have some Gauss Law effects. Assume a random (even) distribution of electrons in a sphere. If a test electron (remember you always have to keep your frame of reference in mind) is outside of all the electrons, then it will be repelled by all of the other electrons and it will fly away and hit the wall with as much acceleration that is possible. But what about a test electron that is located 1/2 of the way towards the center. It will be repelled by all of the other electrons, but some of the electrons are in front of it, behind it, behind and to the left, etc, etc. It will experience net acceleration only towards the area where the least amount of electrons oppose it. In that tiny time slice any electrons at a greater radius than it will effectively cancel out each other. You could speak of succeeding shells of electrons that each act as a Gauss sphere to the test electron so long as the test electron is at a smaller radius than the referenced shell. It is a very messy problem to calculate, but if the opposing or attracting charges are fixed to a solid surface that is spherical and that is at a greater radius than the test charge you have the classic example that is used to derive Gauss Law in simple terms. This does not preclude other free charged particles within the sphere interacting with the test particle. But since the issue is what effects the charged particles (or unit of surface area) fixed to the sphere has, the answer is that they have no effect, they cancel out.

As far as electrical currents arcing between shells, so what.. A sphere within a sphere. The inner sphere will will not see the charge on the outer sphere, but the outer sphere will see the charge on the inner sphere, as it is essentially the same as a point charge. T he current will flow around it on it's way to the ground it is seeking . The driving voltage is not betwenn the spheres, but from the outer sphere to to ground. If the inner sphere gets in the way it will conduct the charge through its metal walls. The Faraday cage protecting you from a lightning strike is the same principle ( I think). A current may develop inside the outer sphere for what ever reason, but it does effect the inner sphere in an ideal situation (no stalks, no holes that are too big, etc.). Inside the Faraday cage and wearing your metal suit, a sudden intense current flow does not induce a charge in you (or your suit) that accelerates you and slams you against the inside surface of the Faraday cage. The comparison with stacked transformers is similar. The voltage between the ends may be tremendous, but it can be set up so that the transformers see only the voltage on neighboring transformers.

And actually, I think that if the system is ideal with perfectly smooth spherical, and conductive surfaces, and no stalks even if the internal sphere is off center there would be no potential difference that would build up on the internal sphere relative to the outer sphere, so you could apply any voltage to the outer sphere and there would be no current flow. Of course, if you could apply a voltage to the inner sphere (that would be difficult to do without compromising the ideal situation) current could flow to the outer grid provided there was a charge carrier available (free ions and/ or electrons). There is nothing in Gauss Law that prevents this.

This is very close to what happens with electrons within the Polywell Magrid. They repel themselves towards the Magrid. The magnetic shielding impedes (turns) this but does not modify the electrostatic causes for this action. But the positively charged Magrid does not attract the the internal electrons, they are driven only by the net negative charge that is concentrated more centrally than the test electron. Also, note that these are ideal considerations. There are holes and this compromises this situation, I have have no idea of the magnitude of these deviations from the ideal, but it must not be dominate.

Consider the injection and recirculation of electrons. There is a resistance towards central convergence due to the space charge within the Wiffleball. To overcome this the electrons need to be accelerated towards the center with an external force (charge on the Magrid). IF the electron saw the Magrid potential while within the Magrid then when the electron reached the Wiffleball border or a cusp it would not have a KE of ~ 10,000 eV, but this plus the Magrid voltage for a total of 22,000 eV. Upon exit the electron would be flying away at 22,000 eV. The Magrid positive charge would now decelerate it to ~ 10,000eV of outward directed KE. There would be no possibility of recirculation. So any discussion of recirculation is ridiculous unless the Gauss Law considerations is at least approximately relevant. If you do not consider the effects of Gauss Law in you simulation, of course there will be no recirculation.

Dan Tibbets

[KitemanSA]

The problem Dan is that you are equating a "spherical conductor with a net charge" to a "uniform sphere of charges". This equation is false.

If you had a NONconductor with a unitform distribution of charges, Gauss' law would apply. But with a conductive sphere, the charges can move due to the presence of an external charge and therefore no longer be uniformly distributed.

In your gravitational analog, a conductive sphere with charge would be like having a massless rigid spherical shell with liquid on it. If you bring a mass near such a surface from the inside (or outside for that matter) the liguid will gravitate toward the mass and the gravitational field will no longer follow Gauss' law.

At least that is what I think they are trying to get at.