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How do you make the potential zero at infinity, here

  1. May 11, 2009 #1
    Potential difference is what which makes sense. But calculating the potential difference b/w two points and one of them being infinity, where we define potential to be zero, we can actually find the potential at the point. Fine, agreed. Now, how do you define the potential at infinity to be zero. E.g look at this thread: https://www.physicsforums.com/showthread.php?t=79683

    The following point made in the post (by Andrew Mason) puzzles me.

    http://www.flickr.com/photos/37453425@N07/3522609499/in/photostream/

    Apparently, he has put ln(infinity) to be zero to get the solution. I feel that this isn't the right way to define the potential at infinity be zero. ('cos of course ln(infinity) is not zero) The solution, which is right, is also used in Feynman Lectures Vol II Section 14-3.

    Am I missing out something here?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 11, 2009 #2
  4. May 12, 2009 #3

    rcgldr

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    The formula in the image is wrong. Link to proper formula and integral:

    https://www.physicsforums.com/showthread.php?t=79189

    The potential energy is based on the work done by "moving" from infinity to some point "r".

    [tex]
    F = -\frac{GMm}{r^2}
    [/tex]

    [tex]
    U = -\int_\infty^r \frac{GMm}{s^2} ds
    [/tex]

    [tex]
    U = GMm(\frac{1}{r} - \frac{1}{\infty}) = GMm(\frac{1}{r})
    [/tex]
     
    Last edited: May 12, 2009
  5. May 12, 2009 #4
    The formula in the image is for an infinitely long charged cylinder. The formula given is right (At least Feynman says so! - Refer the section I've mentioned earlier)

    My question is: In deriving the formula for the potential, do you put ln(infinity) = 0 as in the linked thread (the part is shown in the image)
     
  6. May 12, 2009 #5

    Born2bwire

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    I would say that the given formula is incorrect. He is calculating the potential difference, which would be infinite. The potential is the spatial integral of the electric field evaluated at a given point. The potential difference is the spatial integral of the electric field along a given path. With a 1/r potential, it will take an infinite amount of energy to move the charge from infinite to a given finite distance r.

    For the most part though, potential is ambiguous, it is only really useful if we have a point of reference. In the question of his capacitance problem, the voltage difference between a point and infinite doesn't seem relevant. They are asking the capacitance of the cylinder due to the induction of charge from the line source. The cylinder is held to ground, so the surface of the cylinder should be given as having a potential of zero. Then your potential difference should be taken from the surface of the cylinder with the potential on the cylinder as being zero.

    So I think Andrew may have made a mistake in defining his zero potential reference point.
     
    Last edited: May 12, 2009
  7. May 12, 2009 #6
    Absolutely.
    All my doubt is: If you refer this - "Feynman Lectures Vol II Section 14-3" - he has written the formula for Potential (difference with the reference at infinity) arising due to infinitely long charged cylinder. Jus' show me the derivation of that.
     
  8. May 12, 2009 #7

    Born2bwire

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    What does Feynman say? I don't have a copy the text at my office and I already made my trip to the library today to get textbooks on special relativity (bedtime reading, joy!).
     
  9. May 12, 2009 #8

    rcgldr

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    In general, if the attractive force from a point is relative to 1/r2 then for an infinitely long line it's 1/r and for an infinitely large plane, it's constant.

    For for the cylinder case the standard definition doesn't work:

    [tex]
    U = -c\int_\infty^r \frac{1}{s} ds
    [/tex]

    [tex]
    U = -c({ln}(r) - {ln}(\infty))
    [/tex]

    For the cylinder case, potential energy could be redefined as:

    [tex]
    U = c\int_1^r \frac{1}{s} ds
    [/tex]

    [tex]
    U = c({ln}(r) - {ln}(1)) = c\ {ln}(r)
    [/tex]

    For the plane case, potential energy could be redefined as:

    [tex]
    U = c\int_0^r \ ds
    [/tex]

    [tex]
    U = c(r - 0) = c\ r
    [/tex]
     
    Last edited: May 12, 2009
  10. May 12, 2009 #9
    Oh... so you change the reference... Thnx :)

    btw, this means the thing done in the post, I gave link to, is wrong... rite?
     
  11. May 12, 2009 #10

    rcgldr

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    Yes, [itex]{ln}(\infty) = \infty[/itex].
     
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