Mastering Physics: Electron turning 90 degrees in capacitor

AI Thread Summary
To determine the electric field strength needed for an electron to turn 90 degrees in a parallel-plate capacitor, the vertical velocity must be brought to zero while maintaining horizontal velocity. The calculations involve using kinetic energy and the relationship between force, mass, and electric field. Initial attempts yielded an electric field value of 17655 N/C, but corrections were needed regarding the acceleration components and factors in the equations. After addressing these issues, the correct total magnitude of the electric field was found using the Pythagorean theorem. The final solution was confirmed as accurate following these adjustments.
danielhep
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Homework Statement


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?
upload_2017-3-11_15-4-48.png


Homework Equations


Vf2=Vi2+2ad
F=ma
F=Eq
K (kinetic energy) = 1/2mv2

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.

Vi=(2K/m)1/2
d=.01*cos45
0=(2K/m)+2a*.01*cos45
-K/(m(.01*cos45))=a

Now finding acceleration in terms of E
qE=ma
qE/m=a

Combining them:
-K/(m(.01*cos45))=qE/m
-K/(q(.01*cos45))=E

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!
 
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danielhep said:
Vi=(2K/m)1/2
d=.01*cos45
0=(2K/m)+.01*cos45
Did you leave out the acceleration in the second equation above?
[EDIT: Should there also be a factor of 2 in the second term?]
-K/(m(.01*cos45))=a
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?
 
TSny said:
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?
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.
 
Good work!
 
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