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Homework Help: Mastering Physics: Electron turning 90 degrees in capacitor

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A problem of practical interest is to make a beam of electrons turn a 90∘ corner. This can be done with the parallel-plate capacitor shown in the figure (Figure 1) . An electron with kinetic energy 2.0×10−17 J enters through a small hole in the bottom plate of the capacitor.
    What strength electric field is needed if the electron is to emerge from an exit hole 1.0 cm away from the entrance hole, traveling at right angles to its original direction?

    2. Relevant equations
    K (kinetic energy) = 1/2mv2

    3. The attempt at a solution
    The electric field needs bring vertical velocity to zero and horizontal velocity to what the vertical velocity was. I should just be able to find what E-field is necessary to bring the vertical velocity to zero without having to worry about the horizontal velocity, I think.


    Now finding acceleration in terms of E

    Combining them:

    Now, when I go through this I get 17655 N/C, which seems close (right order of magnitude), but according to Mastering it's not correct. Any help would be appreciated, thank you!
    Last edited: Mar 11, 2017
  2. jcsd
  3. Mar 11, 2017 #2


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    Did you leave out the acceleration in the second equation above?
    [EDIT: Should there also be a factor of 2 in the second term?]
    What happened to the factor of 2? [EDIT: I think this is ok. See edited comment above]
    Is the acceleration "a" the magnitude of the total acceleration, or does it represent just a component of the acceleration?
  4. Mar 11, 2017 #3
    Fixed the original post, thanks.

    a represents the y component of the acceleration... Which means I'm just finding the y component of the electric field...

    I just took my answer and used the pythagorean theorem to find the total magnitude and it was correct! Thank you! That was a perfect hint.
  5. Mar 11, 2017 #4


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    Good work!
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