Charged particle acceleration across a potential

  • #1

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Hello,
I haven't been able to find the answer to this anywhere.

When calculating the energy gained by a particle that is accelerated across the gap of two cavities (e.g. Dees in a cyclotron, or charged cylindrical cavities of a linear accelerator), does one need to take into account the size of the conductor cavity--or more accurately the distance of the charged particle from the surfaces of these cavities?

If an electron, or proton were simply traveling perpendicular from the face of one charged plate to the face of another plate with a potential difference of, let's say say +10V and -10V (for simplicity sake) in a vacuum, the energy gained, as I understand it, is an increase of 20 electron volts. However, when the particle is inside of (not touching) a charged cylinder or a charged cyclotron Dee at one potential traveling across a gap to the inside of another charged cylinder or cyclotron Dee it is always at some distance from the conductor surface, so it seems like the particle would "feel" less of a potential than if it were going from being in direct contact with one conductor to being in direct contact the other conductor because the potential should drop more the further away it is located from the surface of the conductor. There is a visual representation of a linear accelerator on Wikipedia here: https://en.wikipedia.org/wiki/Linear_particle_accelerator, which may be helpful where you can see the particle traveling down the center of the cylindrical cavities some distance away from the conductor surfaces.
I'm not necessarily looking for a mathematical formula to solve this right now, I just want to know if my thinking is correct that the potential energy gain would actually be somewhat less than E =
electron volts.
Thanks for any help.
Todd
 

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  • #2
RPinPA
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Short answer: yes.

Under the assumption of a constant electric field, the standard assumption between two plates where you are far from the edge, then voltage varies linearly with distance from one plate to the other.

If you have voltage ##\Delta V## across distance ##d##, then the constant electric field is ##E = \Delta V/d##. So across distance ##x < d##, the voltage difference is ##E x = \Delta V (x/d)##. So if it's 20 V from plate to plate and you only go 80% of the distance from plate to plate, the voltage difference is only 80% of 20 V.
 
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  • #3
Thanks for the reply. I drew a simple diagram (attached) for clarity. I'm showing two cavities instead of two plates per my previous question. I assume the acceleration of the particle would be determined the same way, and by the same equation you provided, but one would have to determine the potential at the "start" and "end" positions of the particle which for this exercise we'll say are at the center of the inside edges of the cylinders. I would assume that the potential at these start and end points are less than the potential at the surfaces of the cylinders, and therefore the acceleration would be less than 20 electron volts. Is this correct?
Thanks,
Todd
 

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  • #4
RPinPA
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Thanks for the reply. I drew a simple diagram (attached) for clarity. I'm showing two cavities instead of two plates per my previous question. I assume the acceleration of the particle would be determined the same way, and by the same equation you provided, but one would have to determine the potential at the "start" and "end" positions of the particle which for this exercise we'll say are at the center of the inside edges of the cylinders. I would assume that the potential at these start and end points are less than the potential at the surfaces of the cylinders, and therefore the acceleration would be less than 20 electron volts. Is this correct?
Thanks,
Todd
The depicted configuration is very different from a parallel plate capacitor. The electric field is more complicated than a constant, and I would hesitate to conclude that the entire interior of the +10V cylinder is at +10V. So don't just blindly apply parallel-plate or constant-field equations to a totally different problem.

You'd have to do the calculation, would would be something similar to what is done in this thread.
https://www.physicsforums.com/threads/electric-potential-of-a-hollow-cylinder.582276/

Here's a little more analysis of this specific configuration
https://physics.stackexchange.com/questions/92123/why-doesnt-an-electron-feel-an-electric-field-and-thus-accelerate-whilst-insid
"To prevent that the drift tubes are engineered to arrange a low field region between each accelerating region and ensure that the entire bunch gets a push in the right direction."
OK, so the design is that the electric field is close to zero within the cylinders, and since electric field is the rate of change of voltage, that means the voltage within the cylinders is designed to be constant or nearly so. Thus, it is a reasonable assumption that the voltage is a constant +10V on and inside one cylinder and -10 V for the other.

I'm not sure why in your question the particle is starting and ending away from the cylinders. Where is it before it starts accelerating? Why does it stop in midair? But again the direct answer to your question is that yes, two points which are in the region between the cylinders, will have a voltage difference less than 20 V.
 
  • #5
ZapperZ
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Thanks for the reply. I drew a simple diagram (attached) for clarity. I'm showing two cavities instead of two plates per my previous question. I assume the acceleration of the particle would be determined the same way, and by the same equation you provided, but one would have to determine the potential at the "start" and "end" positions of the particle which for this exercise we'll say are at the center of the inside edges of the cylinders. I would assume that the potential at these start and end points are less than the potential at the surfaces of the cylinders, and therefore the acceleration would be less than 20 electron volts. Is this correct?
Thanks,
Todd
Did you find this configuration from somewhere, or is this something you made up?

Also, are you going somewhere with this? This is not normally what we do in particle accelerators. Most accelerators use RF fields, and these fields are inside waveguides of various geometry, typically cylindrical, i.e. these fields are INSIDE the cylinders, not outside the way you've drawn it. So waveguide physics apply, and this often resulted in fields with, say, TM modes.

Zz.
 
  • #6
Vanadium 50
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Your question seems to boil down to "If there is a potential difference V between point A and point B, what will the potential difference be between two points, one near A and one near B." That can't be answered exactly, but it will be "near" V.

And Zz is right - this is not how accelerators usually work.
 
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  • #7
Thanks RPinPA, Vanandium 50, and ZapperZ,

@ZapperZ, The drawing is something I drew up. I'm toying with the idea of building a small cyclotron using rare earth magnets, so I would be using two hollow Dees that would be alternately charged to produce the acceleration. I was curious if the acceleration of a particle across the gap in the Dees expressed in eV would be easily calculated using the difference in potential of the two Dees, or if the potential difference across the gap is actually less where the particle's path actually is--away from the surfaces of the conducting surfaces of the Dee. It seems pretty clear from all of everyone's comments that it is less, but that it gets complicated to calculate the field. I felt like that should be true, but it seems like whenever I read about cyclotrons, or old style electrostatic linear accelerators the description of how it works is always simplified to seem like the exact gain in energy is more easily calculated than it really is. I'm guessing a computer simulation might be a good way to go if I really wanted to know. The beauty of the cyclotron is that it doesn't really matter that much, the particle just needs to go around enough times to get up to speed.

I had wondered about this for a while, but I got interested in the question again when was reading about electrostatic lenses (which I may play around with also), like an Einzel lens. There are some similar dynamics going on there.
 
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berkeman
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I'm toying with the idea of building a small cyclotron using rare earth magnets,
Thread closed for Moderation...

After review, the thread will stay closed. Building even a small accelerator involves dangerous activities like dealing with x-rays, high voltages and high vacuum levels. Please seek out a local Mentor to help you with such projects.
 
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