Timeline for How does electron gun accelerates electrons?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 27, 2011 at 9:44 | comment | added | jartza | @dmckee: this articles says that no, you can not use lined up charged capacitors as a particle accelerator: link | |
| Jul 26, 2011 at 23:35 | comment | added | dmckee --- ex-moderator kitten | I know how you came by this answer...you're thinking a charge $e$ gets energy $V\text{ electron volts}$ when accelerated through a potential difference of $V\text{ volts}$, and you're trying to apply it to this case assuming that infinite distance represents a well defined base potential, but go back to a more basic definition. Use $W = \int q\vec{E} \cdot d\vec{l}$ and think about the field. | |
| Jul 26, 2011 at 23:19 | comment | added | dmckee --- ex-moderator kitten | Here is an example (this one does not use the parallel plate geometry). Notice that the emitter is located in a region of high field? That's important for the initial pick-off | |
| Jul 26, 2011 at 23:16 | comment | added | dmckee --- ex-moderator kitten | @jartza: No. In the limit of a the plate dimensions much greater than the separation nothing outside the thing. Moreover, a typical design for the initial gun does not start the emitter "far away", but near the cathode inside the field region. In particle physics we can keep throwing energy into electrons into the tens of giga-electron volts and send them swooping round accelerator rings miles in length. If your understanding were correct none of that would be possible. | |
| Jul 26, 2011 at 22:50 | comment | added | jartza | @dmckee: many meters of weak field outside, a few millimeters of strong field inside | |
| Jul 26, 2011 at 20:54 | comment | added | dmckee --- ex-moderator kitten | @jartza: You answer is physically wrong even then because for roughly planar electrodes there is little field outside the electrodes (zero in the ideal limit). | |
| Jul 26, 2011 at 20:16 | comment | added | jartza | @Ben Crowell Electron that is shot from a far away position towards the negative plate of a parallel plate capacitor, loses 3 electron-volts of energy when traveling towards the negative plate, gains 6 electron-volts of energy when traveling between the plates, and loses 3 electron-volts of energy when traveling away from the positive plate. Energy loss must be equal to energy gain, and it must be symmetric. So exactly half of gained energy is lost when electron travels from positive plate to infinity. | |
| Jul 26, 2011 at 16:54 | comment | added | user4552 | @dmckee: He says "Electron that is shot from a far away position...," which means that his claim isn't that the gun doesn't work. It's certainly true that when an electron is brought from infinity into some electrostatic device and them removed again to infinity, it ends up with the same energy it started with. It's also irrelevant. And I think the 1/2 is simply due to some conceptual confusion, such as a possible belief that when an electron is outside the plates it only feels the field from the nearer plate. | |
| Jul 26, 2011 at 16:49 | comment | added | dmckee --- ex-moderator kitten | Your claim here amounts to "the gun doesn't work because the electron loses the same amount of energy as it gains.", which is nonsense. Perhaps you should look for a reference to back this up. | |
| Jul 26, 2011 at 16:19 | comment | added | Georg | "The positive grid" means he knows nothing, There are several positive grids in a CRT. | |
| Jul 26, 2011 at 16:05 | comment | added | user4552 | Where do you get this figure of 1/2? It doesn't sound right to me. Are you claiming that it's exact? Approximate? | |
| Jul 26, 2011 at 16:04 | comment | added | Georg | I deleted my answer, because I do not want to compete with such nonsense. | |
| Jul 26, 2011 at 15:34 | vote | accept | BarsMonster | ||
| Jul 27, 2011 at 22:59 | |||||
| Jul 26, 2011 at 15:15 | history | answered | jartza | CC BY-SA 3.0 |