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Homework Help: Electrical potential difference given velocity

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    An electron moving parallel to the x axis has in initial speed of 3.7x10^6m/s at the origin. Its speed is reduced to 1.4x10^5m/s at the point x=2.0 cm. What is the potential difference between the origin and this point? Which point is at the higher potential?

    2. Relevant equations
    I am not sure what equations are useful with this equation.

    3. The attempt at a solution
    I have not been able to attempt this problem, I have no idea how i would go about starting it
  2. jcsd
  3. Sep 16, 2009 #2
    I'm a little unsure, but perhaps you could use the fact that Uq = 0,5 m v^2, at least for the first questions. I don't quite know what is meant by the second question, but electrons are negative. So try that.
  4. Sep 16, 2009 #3
    Anden, examine the equation you posted carefully, what happens if the charge is negative? Does that make sense physically?

    What is the electron's kinetic energy at the origin?
    What is its kinetic energy at point x?

    What does this say about the work done on the electron?

    What is the relationship between the work done on a mass moving it from point A to point B, and the potential difference between those two points?
  5. Sep 16, 2009 #4
    Is it really that simple? there is nothing special about it being in the electrical potential section?

    If its Uq then is q the charge of the electron? Or do i simply have to find the kinetic energy?
  6. Sep 16, 2009 #5
    Ups, yes of course you're right, the electron has negative charge. But I think if you forget that and count the electrons charge as positive, you can use the equation to get directly to the potential difference.

    EDIT: A thought: If you get a negative potential difference, doesn't that describe whats going on here? I mean, you get a general idea of the direction of the acceleration of the electron.
  7. Sep 16, 2009 #6
    Oh, so since it comes out nonsensical (A square equaling a negative) we just ignore the problem? :P

    I haven't brushed up on the subject in a while, but I'm pretty sure that [tex]U_{AB}\cdot q=W_{AB}[/tex]
  8. Sep 16, 2009 #7
    What do you mean by a square equaling a negative? And why is it so non-sensical? ;) And what happens with a negative charge in Uq = W?
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