Is the Setup for the Electric Scaler Potential Integral Correct?

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Discussion Overview

The discussion revolves around the setup of an electric potential integral in the context of cylindrical coordinates. Participants are examining the mathematical formulation and dimensional consistency of the integral related to surface charge density and the distance from an observation point to the surface charge.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant defines the distance R' as R'=√((-r)²+(-Φ)²+(z)²) and expresses confusion about the complexity of the integral setup.
  • Another participant questions the definition of R' and seeks clarification on its formulation.
  • A participant explains the use of cylindrical coordinates, stating that the subtraction leads to the vector R' and its magnitude.
  • Concerns are raised about dimensional consistency, with a participant noting that adding terms with different dimensions (like rad² and m²) does not make sense.
  • There is a consideration of whether the distance R' varies with different angles Φ, leading to a realization that it does not, as it depends only on r and z.
  • Participants discuss the appropriate differential area for the surface integral, debating whether it should be drdΦ or dΦdz, with a suggestion that drdΦ is more appropriate based on the radial and angular definitions.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correctness of the initial setup and the dimensional consistency of the integral. There is no consensus on the final formulation of the integral or the correct differential area for the surface integral.

Contextual Notes

The discussion highlights potential limitations in the initial assumptions about the definitions and dimensions involved in the integral setup. The dependence on the choice of coordinates and the nature of the surface integral remains unresolved.

kosmocomet
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Homework Statement


66e3b29a-566a-4e53-b5cb-82d593a806dd
upload_2018-4-1_10-10-18.png


Homework Equations


  • V=¼*(1/(π∈) * ∫(ρs/(R')*ds' where R' is distance from point to surface
  • R'=|R-Ri| distance from observation point to location of surface charge density.

The Attempt at a Solution


So my attempt was to define R' as R'=√((-r)2+(-Φ)2+(z)2). Then I said that ds' is = rdrdΦ. I put this into the integral. I am confused since the integral looks very involved ad wanted to know if my set up was correct.

Any help is much appreciated.
 

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kosmocomet said:

Homework Statement


66e3b29a-566a-4e53-b5cb-82d593a806dd
View attachment 223189

Homework Equations


  • V=¼*(1/(π∈) * ∫(ρs/(R')*ds' where R' is distance from point to surface
  • R'=|R-Ri| distance from observation point to location of surface charge density.

The Attempt at a Solution


So my attempt was to define R' as R'=√((-r)2+(-Φ)2+(z)2). Then I said that ds' is = rdrdΦ. I put this into the integral. I am confused since the integral looks very involved ad wanted to know if my set up was correct.

Any help is much appreciated.
Why do you define R' that way?
 
tnich said:
Why do you define R' that way?
Thanks for the response. I am using cylindrical coordinates and since the point P is (0,0,z) and the disk is(r,Φ,0) doing the substraction is -r,-Φ,z whcih is the vector R'. The magnitude is then R'=√((-r)2+(-Φ)2+(z)2).
 
kosmocomet said:
Thanks for the response. I am using cylindrical coordinates and since the point P is (0,0,z) and the disk is(r,Φ,0) doing the substraction is -r,-Φ,z whcih is the vector R'. The magnitude is then R'=√((-r)2+(-Φ)2+(z)2).
OK. I see how you did it, but it is not correct. For one thing it is dimensionally inconsistent. It does not make sense to add ##rad^2## to ##m^2##. Let's approach it another way. Let r and z be constant and consider two angles, ##Φ_1 = 0## and ##Φ_2 = π/2##. Is the distance R' different for the different values of ##Φ##?
 
Well...I guess not mow that I think about it. Since, the distance would be based on r and z, correct?
 
If this is true, would the surface integral be drdΦ or dΦdz? I think it would be drdΦ since it is defined in radial and angular direction. The surface, I mean.
 
kosmocomet said:
Well...I guess not mow that I think about it. Since, the distance would be based on r and z, correct?
Yes.
 
kosmocomet said:
If this is true, would the surface integral be drdΦ or dΦdz? I think it would be drdΦ since it is defined in radial and angular direction. The surface, I mean.
That's close, but what is the differential area in polar coordinates?
 

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