Electric Field of a Hollow Cylidrical Conductor

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Homework Help Overview

The discussion revolves around determining the electric field of a hollow cylindrical conductor with a positive surface charge density in a vacuum. The problem involves applying Gauss' law to find the electric field's magnitude both inside and outside the conductor.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate Gaussian surface to use, considering both cylindrical and circular options due to the symmetry of the problem. There are questions about the implications of the conductor being infinite.

Discussion Status

The conversation includes suggestions for using a cylindrical surface to leverage symmetry, and some participants express confidence in their understanding of the approach. However, there is no explicit consensus on the final outcome or solution.

Contextual Notes

Participants are navigating the application of Gauss' law and the implications of the conductor's geometry, with some uncertainty about the correct Gaussian surface to use. The original poster expresses difficulty in progressing with the problem.

Zook104
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Consider a hollow cylindrical conductor in vacuum with its axis aligned with
the z-axis, and with a positive surface charge density σ. The direction of the
electric field is radial and its magnitude E is only a function of the distance r
from the z-axis, E = E(r).

Use Gauss' law to obtain the magnitude of the electric field at r < R and
at r ≥ R, where R is the radius of the conductor.

I don't know whether I have got my brain stuck in a rut but I can't for the life of me solve this. Any and all help will be greatly appreciated :D
 
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Try using Gauss' law, as it says. What kind of Gaussian surface should you use to take advantage of the symmetry?
 
A cylindrical surface? With surface area of 2\pirl ?
 
Or should I just use a circular one because the conductor is infinite either side?
 
Use the cylinder with surface area of 2*PI*r*l
Do the calculation using the information they have given you, and it should all come out nicely. (Even though the conductor is infinite, it doesn't go bad).
 
Zook104 said:
A cylindrical surface? With surface area of 2\pirl ?

Go with that one.
 
Excellent I believe I have got the correct answer :D Thank you so much for your help
 

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