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Question about symmetry with potential fields.

  1. Apr 28, 2012 #1
    Quick question (I think anyway)

    I'm currently trying to solve a problem that essentially is asking for me to find the final velocity of an electron that is travelling between a cathode and an anode of potential difference 300. At half the distance to the anode can I assume by symmetry that it would be half the total potential difference i.e. 150V?
     
  2. jcsd
  3. Apr 28, 2012 #2

    gneill

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    Staff: Mentor

    You could... How do think that will help you?
     
  4. Apr 29, 2012 #3
    Well if I know the potential at one end is say, 300V and I arbitrarily set the cathode side to 0 for sake of convenience -knowing that it is only the difference in potential that is important- and I want two values; one half way and one full displacement, symmetry can help a lot. If I make that argument I can conclude that half the distance would be half the potential.
     
  5. Apr 29, 2012 #4

    gneill

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    Oookaaay... still not seeing the utility of that for solving a problem which is otherwise quite straightforward. However, I am prepared to be amazed :smile:
     
  6. Apr 29, 2012 #5
    I was attempting to try it with conservation of energy.

    [tex]
    \frac{1}{2} m(v^2)_i + q\phi_i = \frac{1}{2} m(v^2)_f + q\phi_f
    [/tex]

    Where [itex]v_i[/itex] is 0, and [itex]\phi_i[/itex] is 300 and q is the charge of the electron. Manipulation of this equation, isolating final velocity I get a difference of potentials under a square root. That's why I have to make an assumption based on symmetry. Does that work?
     
    Last edited: Apr 29, 2012
  7. Apr 29, 2012 #6

    gneill

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    If I may suggest, why not calculate the work done in moving a charge q through a potential difference V?
     
  8. Apr 29, 2012 #7
    Hmmm, seemed to have overlooked such an obvious way of doing it. I will get right on that. In any case does my argument have any validity?

    Thanks for the pointers by the way.
     
  9. Apr 29, 2012 #8

    gneill

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    I'm afraid I just can't picture a way forward with it; to me it just seems to divide the problem into two problems with the same unknown quantities.
     
  10. Apr 29, 2012 #9
    I don't know. It only leaves one unknown being [itex] \phi_f [/itex] with the obvious exception of the final velocity. Which brings me back to the initial question, can I assume the potential would be half its max (or min) at half the distance? If it is, then I only have the one unknown of the final velocity. Maybe I'm overlooking something, but it seems pretty easy to solve that way as well so long as I can make that assumption with respect to the potential.
     
  11. Apr 29, 2012 #10

    gneill

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    The electric field between (sufficiently large) plates is essentially uniform, so yes, halfway between the plates the potential should be half as well.
     
  12. Apr 29, 2012 #11
    Great, thanks.
     
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