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Question Pertaining to Voltage

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    If an electron travels 0.350 m from an electron gun to a TV screen in 46.0 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)



    2. Relevant equations

    E=k(q/r^2) ΔV=|Ed| ΔKE + ΔPE = 0

    3. The attempt at a solution

    I have no clue how to answer this question. I don't know if it's because I am trying to use the wrong equations, or because I am just not seeing the connections
     
  2. jcsd
  3. Feb 10, 2013 #2

    rock.freak667

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    Homework Helper

    If the electron travels 0.35 m in 46 ns, what velocity does it travel at? Hence what KE does it posses at this velocity? Also how does voltage and charge relate to energy?
     
  4. Feb 10, 2013 #3
    If the electron travels 0.35 m in 46 ns, what velocity does it travel at? Hence what KE does it posses at this velocity? Also how does voltage and charge relate to energy?

    Well the velocity is going to be at about 7.6x10^6 m/s. I know that K.E.= 1/2mv^2, so can I calculate the kinetic energy of that and use it as the energy to calculate voltage? I am a little lost
     
  5. Feb 10, 2013 #4

    gneill

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    Staff: Mentor

    Keep in mind that the electron will be accelerating throughout its journey; It doesn't cover the interval at a constant speed.

    What's the formula for the energy gained by a charged particle falling through a given potential difference?
     
  6. Feb 10, 2013 #5
    E = |qΔV|? is that the equation you are talking about?
     
  7. Feb 10, 2013 #6

    gneill

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    Staff: Mentor

    Yes.
     
  8. Feb 10, 2013 #7
    Well the only thing that I can think of using is the Δ Kinetic Energy + Δ Electric Potential=0 and then trying to solve for V (voltage).
     
  9. Feb 10, 2013 #8

    gneill

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    Sounds like a good plan :wink:
     
  10. Feb 10, 2013 #9
    Ha ha okay, I actually tried doing that earlier but I got the wrong answer. Which of the Velocities and voltages are zero? And for the mass of the electron do I use 9.1X10^-31?
     
  11. Feb 10, 2013 #10

    gneill

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    Staff: Mentor

    Show your work so we can see where something might have gone wrong.

    The electron starts out with zero velocity and accelerates to the screen thanks to the potential difference.
     
  12. Feb 10, 2013 #11
    1.)1/2m(Vf^2-Vi^2) + q(Vf-Vi)=0, cancel out initial velocity and final voltage (I think) since they are zero.
    2.)1/2mVf^2-qV=0
    3.) 1/2mVf^2=qV
    4.)1/2mVf^2/q=V

    That's what I tried, and got the wrong answer so I guess that I am setting up the equation wrong.
     
  13. Feb 10, 2013 #12
    m=mass of electron, 9.1x10^-31 kg
    Vf= 7.61x10^6
    q= 1.6x10^-19

    Those are the numbers I am using
     
  14. Feb 11, 2013 #13

    gneill

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    How did you determine Vf? Remember that the electron is undergoing accelerated motion, not constant velocity...
     
  15. Feb 11, 2013 #14
    Hmmm that is true. I am kinda stumped then
     
  16. Feb 11, 2013 #15
    How did you get this Velocity ?

    I think you have erred in calculating the Final Velocity at x=0.35m
     
  17. Feb 11, 2013 #16

    gneill

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    Staff: Mentor

    It's just the usual accelerated motion of a projectile. Use the information given (distance and time) to find the final velocity assuming that a constant acceleration is applied. Hint: you can begin by determining the acceleration.
     
  18. Feb 11, 2013 #17
  19. Feb 11, 2013 #18
    Okay, I will try to figure it out. In the meantime, can you help me with another question?


    The problem is as follows:


    In nuclear fission, a nucleus splits roughly in half.

    (a) What is the potential 3.50 10-14 m from a fragment that has 50 protons in it?
    I calculated this to be 2.05x10^6, which is correct.


    (b) What is the potential energy in MeV of a similarly charged fragment at this distance?

    I have no idea what this part is asking for; our professor did not go over those units at all and the textbook we are using is very vague on the subject.
     
  20. Feb 11, 2013 #19

    gneill

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    Staff: Mentor

    eV stands for 'electron volt'. It's a unit of energy equivalent to the energy gained by a unit electric charge (the magnitude of the charge on the electron) falling through a potential difference of 1 volt. Hence electron volt.

    MeV is Mega eV, that is, a million eV.
     
  21. Feb 11, 2013 #20
    Okay, so how can I go from the volts to ev?
     
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