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Finding an kinetic energy of an electron moving to a positive plate

  1. Jul 12, 2011 #1
    1. The problem statement, all variables and given/known data
    2 part problem. I've already solved the first part.

    Two parallel plates are held at potentials of +380.0 V and -150.0 V. The plates are separated by 3.00 cm.
    a. Find the electric field between the plates.
    b. An electron is initially placed halfway between the plates. Find its kinetic energy when it hits the positive plate.

    2. Relevant equations
    E = deltaV/d
    W = delta U = qEd


    3. The attempt at a solution

    For the first one I plugged into delta V, which was simply 380 - -150 = 530 V.
    then took 530V/(0.03m) = 17,666 V/m
    the 2nd part isn't working.
    q(E)d = (1.6e-19 C)(17,666 V/m)(0.015 m)
    = 4.2398 e-17 CV (needs to be in units of eV?)

    This answer is wrong. Anyone know what I'm doing it wrong?

    by the way. I think it is obvious I'm using the conservation of energy. All potential energy at the start, converts to all kinetic energy at the end.
     
  2. jcsd
  3. Jul 12, 2011 #2

    rock.freak667

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    You might just need to do ΔW = ΔV*q, since the electric field is uniform.
     
  4. Jul 12, 2011 #3
    I tried that way too. Didn't work. deltV = 530 and q= 1.60e-19. Multiply the two and it's equal to 8.49e10-17. Still saying it's wrong.
     
  5. Jul 12, 2011 #4

    Redbelly98

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    Is that how you wrote the answer? You really mean 8.49e-17 here. Also, do they require you to include units in your answer? And how many significant figures should your answer have?

    You basically have it right.
     
  6. Jul 12, 2011 #5
    Sorry. I wrote it exactly as 8.49e-17 eV. Should have 3 sig figs. The practice problems which are the same problem but slightly different numbers have answers like 205 eV when the numbers change to "Two parallel plates are held at potentials of +230.0 V and -180.0 V. The plates are separated by 8.00 cm. "
     
  7. Jul 12, 2011 #6

    Redbelly98

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    Okay.

    But the voltages are given to an accuracy of 0.1V. I.e., how many sig figs are in "380.0 V"?

    Oh. In that case, you might also try answering in units of eV.
     
  8. Jul 12, 2011 #7
    8.49e-17 is in units of CV? How do I convert that to eV?
     
  9. Jul 12, 2011 #8

    Redbelly98

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    CV units are equivalent to Joules (J). Can you convert between Joules and eV?
     
  10. Jul 12, 2011 #9

    rock.freak667

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    Firstly when writing the units of energy (any kind) it's best to write as Joule and not CV. It is correct yes but less confusing that just J.

    Second thing, 1 eV = 1.6(10-19) C * 1 V = 1.6(10-19) J

    so just divide your answer by 1.6(10-19) and it will be in eV.
     
  11. Jul 12, 2011 #10
    ahhh. I had no idea CV = J. Yeah, I knew I could simply divide J/1.6e-19 to get eV. That would make the answer 530.62 eV. This makes sense, but the answer was also marked as wrong : /

    I'm not really sure what I'm doing wrong. Anymore ideas?
     
  12. Jul 12, 2011 #11

    Redbelly98

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    530.62 ... that's remarkably close to the number you got for question (a).

    Think about it: you had 530 (really 530.0) for the voltage between the plates. You multiplied that by 1.6e-19, and then later you divided it by the same number, 1.6e-19.

    Again, you mainly have it right. The problem is with sig figs, or units, or it's always possible the website has it wrong too.
     
  13. Jul 12, 2011 #12
    Thanks. I will email my teacher and report back as to what's going on.
     
  14. Jul 12, 2011 #13

    gneill

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    The electron is placed halfway between the plates. So the ΔV that it falls through is half of 530V. An electron falling through 1V is attributed 1eV. So an electron falling through 530/2 Volts must acquire..._____ eV ?
     
  15. Jul 12, 2011 #14
    Is it really that easy? I hope that wasn't my only mistake. I've since emailed my teacher and asked for another submission. I'll see if that's right. I imagine it would be.
     
  16. Jul 13, 2011 #15

    Redbelly98

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    Ah, can't believe I missed that. :redface:
     
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