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Homework Help: Cahnge in potential difference of moving electron

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    As an electron moves through a region of space, its speed increases from 2.2e^6 m/s to 6.0e^6 m/s. The electric force is the only force acting on the electron.

    a) Did the electron move to a higher potential or a lower potential?

    b)Across what potential difference did the electron travel?

    2. Relevant equations

    well for part b, i'm not even sure what to use.

    K = 1/2mv^2

    U=q x V

    3. The attempt at a solution

    Well i know part a), the electron moves to a higher potential.

    I really don't know how to solve this since the distance or time is not given. And I may be confused as to the difference between potential difference and electric potential. I don't know where to start!
  2. jcsd
  3. Sep 26, 2009 #2


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    Homework Helper

    a) Electron moves in the apposite direction to the electric field. In the electric field the potential decreases along its direction. Then...?
    b) The electron accelerates when it moves through lower potential to the higher potential. Hence its kinetic energy increases. Use the relevant equations and find the potential difference.
  4. Sep 26, 2009 #3
    Yes, i understand the electron is accelerating. obviously, its speed increases, so its kinetic energy also increases. I understand all these concepts.

    I don't know which equations to use. That is where I'm stuck. All the equations I can think of using velocity have time or distance variable, which i can't use.

    The only one I know is K = 1/2mv^2

    So would I do [tex]\Delta[/tex]K = 1/2mv2^2 - 1/2mv1^2

    and since [tex]\Delta[/tex]K = - [tex]\Delta[/tex]UE

    I can then use [tex]\Delta[/tex]V = [tex]\Delta[/tex]U/e to get the potential difference???

    Is [tex]\Delta[/tex]V potential difference or just potential or electric potential?? that is one area I am confused.

    Thank you
  5. Sep 26, 2009 #4


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    Homework Helper

    Yes. ΔV = ΔU/e. No area is involved.
  6. Sep 26, 2009 #5
    This worked!

    A big thanks to you. Sometimes just trying to explain the problem to someone else helps me solve it anyways.

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