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← 08ps-08 Cathode Ray Tube Solution

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Showing Revision 1 created 10/27/2012 by Amara Bot.

  1. All right. So a good first step for solving this problem is to focus on trying to solve
  2. for the time Δt that the electron spends in this electric field here.
  3. I'm going to use this kinematics equation here.
  4. And you notice that I've just applied it to x and not to y. We'll do the y equation in a second.
  5. And the direction I'm calling x is this direction here. I'm going to call this direction y.
  6. The field is only acting in the y direction
  7. so I know there's not going to be any x acceleration so this term drops out.
  8. And this gives us that the time the electron spends in the field
  9. is equal to the length of the field divided by the speed the electron enters the field with.
  10. All right. So now I'm going to use this kinematics equation to work in the y direction.
  11. And we know that initially the electron enters with only the velocity in the x direction.
  12. This term goes to 0.
  13. We know from our earlier parallel plate problems that the acceleration
  14. along the direction of the field is going to be equal to the strength of the field E
  15. times the charge on the electron divided by the electron's mass.
  16. And this gives us a nice equation for our displacement in the y direction.
  17. Now we can substitute in our value of Δt into this equation,
  18. which gives us this equation here.
  19. And doing some algebra, I can solve for the electric field strength E
  20. in terms of the all these other variables.
  21. Now this looks quite messy but I actually know all these values.
  22. I know the mass of the electron. I know its charge.
  23. And I know the x and y distances and our initial speed.
  24. And plugging in all those values, I get that the electric field strength should be 455 N/C.
  25. That means that if the electric field here is 455 N/C,
  26. we should get this kind of bending and hit this part of the screen.