Distance to stop a charged particle?

[David Kunkle]

Trying to figure out the following scenario:

Grid voltage is 10kV. An ion any distance from the grid is accelerated to exactly 10kV by the time it enters the grid. The ion travels out the other side of the grid and now is accelerated to a stop by the grid at some distance. Question is: At what distance does the ion stop? (Before it reverses and begins to start back toward the grid again of course.)

I would like an equation to figure this for any voltage, or distance and work backward to voltage.

Just spent a lot of time trying to figure this out with google, etc. Now that I realize electric field strength and voltage are not the same thing, and there is no simple way to convert, I give up on my own. Anybody know how to calculate-- or at least a rough approximation?

[prestonbarrows]

Talking about the baseline of a single particle in a static electric potential is very simple.

Lets assume we have a spherical chamber at grounded zero volts surrounding a tiny point of a grid in the center held at -10kV with our power supply and imagine dropping a particle off near the chamber wall. We can forget about integrating forces on the moving particle and all those complicated things. It is just an energy balance problem.

The key for energy balance problems is that energy is conserved. We can transfer energy from one type to another, but the total energy is always the same no matter what. In this case, we have electric potential energy and kinetic energy.

Voltage is electrical potential energy. A charged particle will have the potential energy of

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electricalEnergy = charge * potential

where charge refers to the charge of our particle and potential is the electrical potential at the point in space that the particle is currently located.

The kinetic energy is related to speed,

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kineticEnergy = 0.5 * mass * velocity^2

Here the mass is the mass of our ion.

Again, our total energy is always fixed so we can just sum everything up and know that it will always be constant. Then we can figure out all the useful things like the maximum velocity and where the particle will turn around. So I'll just sum up the energies as described above and set them equal to some constant.

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q*phi + 0.5*m*v^2 = constant

OK so what is this constant? We can plug in the values we know to find out. When we let go of our ion, it has a speed of zero and is located at ground where the potential phi is also zero; so our constant is zero in this case. We will see that the exact value does not really matter though.

That is basically it! At any point in time our equation ALWAYS equals zero; if we move to a lower potential we must be moving faster and so on. If you want to know how fast the particle is moving at any location, say the center, plug in the potential at that location and our constant and solve for velocity.

For your question about where it would turn around, we know that the instant something reverses directions its velocity is zero. So we can plug v=0 into our equation and find that this happens when the potential is also zero. For our perfectly spherical case this means that the particle would turn around at the same exact distance away as we let it go. Even without symmetry, it will always turn around when it reaches a location with the same potential as where you first let it go.

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All of this is exactly analogous to gravitational potential energy. In that case, the potential is related to height rather than voltage. So if you have a frictionless bowl and let go of a marble near the edge, it will speed up as it falls towards the center and turn around on the other side at the same exact height as you first let it go.

Things become much more complicated in the real world where a plasma is made of an electrically charged fluid of colliding particles which can also effect the background fields. The full description is well outside the scope of a forum post, or even a few terms of courses.

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Now that I realize electric field strength and voltage are not the same thing, and there is no simple way to convert

Electric field strength is just how quickly the voltage (potential) changes over distance. Look at the units, potential is measured in volts and electric field is measured in volts per meter. So if you have a plate at zero volts and another plate at 10kV and they are spaced 1 meter apart, the electric field in the space between them will be 10 kv/m ideally. The math gets messy around the edges of the plates, but we can just assume that the plates are really big so that the edges are far away and not worry about them for now.

If you are familiar with calculus, the electric field is the negative gradient of the potential.

In our gravity example, this is like the slope of the side of the bowl.

Does that make sense?

[Rich Feldman]

Grid voltage is -10 kV with respect to what?
What is the electric potential at the place where our ion presumably begins to accelerate from rest?

Let's suppose the first answer is "-10 kV with respect to grounded outer shell".
Then the E-field is the gradient of a potential field. Every point inside the shell has a voltage, ranging smoothly from 0 at outer wall to -10e3 at the grid. In the simple case of spherical symmetry, with negligible space charge, the voltage vs radius formula can be found in an E&M textbook discussion of spherical capacitors. http://hyperphysics.phy-astr.gsu.edu/hb ... apsph.html

Now if a neutral atom or molecule becomes singly ionized at a place where the voltage is -7 kV, it will gain only 3 keV in accelerating to the grid.
That's enough to carry it up the potential well on the far side of grid, back to the -7 kV contour.

[David Kunkle]

Thanks for taking the time for the 2 replies. Looks pretty simple now. Whatever distance it started, It'll just "yo-yo" back and forth that same distance each side of the grid.

My problem came from reading more than one post where they stated that under Gauss' Law, a charged particle at rest in an electric field would achieve the same KE or eV by the time it got to the grid - no matter if it started out at 1mm or 1000 KM from the grid. 1 particle could start at 4 cm from the grid and another particle at 8 cm from the grid- yet they would both have the same KE when they got to the grid. This way, downscattered particles get accelerated back to "full" energy. Then, both particles now having the same KE, would stop the same distance out the other side of the grid. That's why I was trying to calculate at what distance they would stop given say a 10kV grid.

Thanks.

[Richard Hull]

Much of this has been dealt with in the past here and a lot of great posts exist discussing it.

Ideal is ideal, theory is theory and reality is a mix of hundreds of mathematical possibilities in the caotic environment of the fusor. There is no math that can give a precise understanding of the action of a fusor. The best we can hope for is possible generalizations based on the physics of what is happening inside.

What is actually happening is a form of controlled bedlam in the fusor with each particle answering all the laws of physics perhaps in a hundered different unpredictable ways during its short lifetime within its little enclosed electrostatic universe. Ultimately, most molecules in a working fusor chamber just get pumped out. This includes fusion debris. Almost none of the Deuterium atoms in the chamber will ever fuse! Almost none in the chamber will even be ionized to become deuterons! A small but decent fraction of the deuterium atoms will wind up buried either permanently or temporarily in the fusor shell's metal lattice.

In a fusor with a measured fusion rate of 2 million fusions per second, there are about 10e15 other particles in the chamber, that is about a dradrillion other particles in that same second, are just roaming around doing nothing! An almost infintestimally, microscopic number of deuterium atoms will fuse purely due to chance rather than by design.

What!?

The design we have is not designed to actually do fusion, but to create an environment where the probability of fusion rises to some small level where it can actually be measured by an adroit and capable scientist armed with good detection gear.

Thus, fusion, as a process, is easy to do here on earth.

Pure fusion is, and always will be, impossible.

Fusion is a probablistic thing and never a given. The best the human race can hope for is to raise the probabilities, by design, to some useful level by a process that requires less energy input than the fusion energy released in that probabilisitc design. So far........ not even close to the goal posts.

Richard Hull

[David Kunkle]

I do get the concept of how chaotic it is inside a plasma. I once saw a diagram of an electron that was somehow tracked thru its course. Tortuous would be a good word to start, and pissed-off bee in a jar might be a good analogy also.

Still wondering about what I read about Gauss's Law. Is that complete hogwash then about particles being accelerated back to uniform energy if they start out at less than 10keV but with a 10kV grid and a grounded chamber?

[Andrew Robinson]

pissed-off bee in a jar

FANTASTIC!

[Richard Hull]

If a particle is ionized negatively at the 7kev gradient point between the grid and shell it will accelerate towards the grid, naturally. However, its ground point, the point of its creation is 7kev and it will gain only 3kev at the grid and orbit about the point of its creation and the grid...... IDEALLY! The fusor is an isolated electrical universe, separate from the one we live in.

The above is why we want all ions to, again, ideally be created at the shell to undergo full acceleration relative to each other at the grid. This will not happen in our heavy (dense) gas environment. Thus, the fusor is all about net loss while at the same time dragging the probability of fusion, kicking and screaming all the way, to a point of detectability to the well instrumented amateur scientist.

Richard Hull