How do you make the potential zero at infinity, here

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Discussion Overview

The discussion revolves around the definition of electric potential at infinity, particularly in the context of calculating potential differences involving an infinitely long charged cylinder. Participants explore the implications of defining potential at infinity as zero and the mathematical reasoning behind it.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of defining potential at infinity as zero, citing that ln(infinity) is not zero and suggesting that this approach may not be correct.
  • Another participant argues that the potential energy is based on the work done moving from infinity to a point, providing a formula for potential energy derived from the gravitational force.
  • Some participants assert that the formula presented in an image is incorrect and relate it to the context of an infinitely long charged cylinder, referencing Feynman's work for clarification.
  • There is a discussion about the ambiguity of potential and the necessity of a reference point, with one participant suggesting that the surface of the cylinder should be considered as having zero potential.
  • One participant expresses a desire for a derivation of the potential formula as presented in Feynman's lectures, indicating a need for further clarification on the topic.
  • Another participant notes that the standard definition of potential energy does not apply uniformly across different geometries, such as point charges, cylinders, and planes, leading to different forms of potential energy expressions.
  • There is a consensus among some participants that the approach taken in the linked post may be incorrect, particularly regarding the treatment of ln(infinity).

Areas of Agreement / Disagreement

Participants express disagreement on the proper treatment of potential at infinity, with multiple competing views on how to define and calculate it. The discussion remains unresolved regarding the correctness of specific formulas and the implications of defining potential at infinity as zero.

Contextual Notes

Participants highlight limitations in the definitions used, particularly regarding the assumptions made about potential at infinity and the dependence on specific geometrical configurations. There are unresolved mathematical steps in the derivations presented.

Shan Ravi
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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?
 
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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".

<br /> F = -\frac{GMm}{r^2}<br />

<br /> U = -\int_\infty^r \frac{GMm}{s^2} ds<br />

<br /> U = GMm(\frac{1}{r} - \frac{1}{\infty}) = GMm(\frac{1}{r})<br />
 
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The formula in the image is wrong.

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)
 
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.
 
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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.
 
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!).
 
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:

<br /> U = -c\int_\infty^r \frac{1}{s} ds<br />

<br /> U = -c({ln}(r) - {ln}(\infty))<br />

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

<br /> U = c\int_1^r \frac{1}{s} ds<br />

<br /> U = c({ln}(r) - {ln}(1)) = c\ {ln}(r)<br />

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

<br /> U = c\int_0^r \ ds<br />

<br /> U = c(r - 0) = c\ r<br />
 
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Oh... so you change the reference... Thnx :)

btw, this means the thing done in the post, I gave link to, is wrong... rite?
 
  • #10
Shan Ravi said:
image from link is wrong
Yes, {ln}(\infty) = \infty.
 

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