How do you make the potential zero at infinity, here

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SUMMARY

The discussion centers on the definition of electric potential at infinity, specifically addressing the assertion made by Andrew Mason that ln(infinity) equals zero. This claim is refuted, as ln(infinity) is indeed infinite, leading to the conclusion that defining potential at infinity as zero is incorrect. The correct approach involves understanding the potential energy derived from the work done moving from infinity to a point r, as outlined in Feynman Lectures Vol II Section 14-3. The potential difference is clarified as the spatial integral of the electric field, emphasizing the importance of a reference point for potential calculations.

PREREQUISITES
  • Understanding of electric potential and potential difference
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of gravitational force and its mathematical representation
  • Basic principles of electrostatics, particularly for charged cylinders
NEXT STEPS
  • Review Feynman Lectures Vol II Section 14-3 for detailed derivation of potential due to charged cylinders
  • Study the concept of electric field and its relation to potential difference
  • Explore the mathematical implications of ln(infinity) in calculus
  • Investigate the potential energy calculations for different geometries, including planes and cylinders
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Students of physics, educators teaching electrostatics, and anyone involved in theoretical physics or electrical engineering who seeks clarity on the concept of electric potential and its reference points.

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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