How Does Gauss's Law Apply to an Infinite Charged Rod?

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Discussion Overview

The discussion centers on the application of Gauss's Law to determine the electric field generated by an infinitely long charged rod. Participants explore the theoretical implications and mathematical reasoning behind using a cylindrical Gaussian surface in this context.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions how Gauss's Law can apply to an infinite rod when the Gaussian cylinder is of finite size, suggesting a potential misunderstanding of the law's implications.
  • Another participant asserts that due to symmetry, the electric field will be purely radial, indicating that the size of the cylinder does not affect the outcome since both the charge enclosed and the length of the cylinder scale linearly.
  • A participant expresses confusion regarding the differing results for finite and infinite rods, noting that the charge enclosed and the area of the Gaussian surface appear to be the same, leading to uncertainty about the electric field's behavior.
  • Some participants explain that charges outside the Gaussian cylinder cancel each other's contributions to the electric field, reinforcing the idea that the field is perpendicular to the rod.
  • There is a discussion about the cancellation of horizontal components of the electric field, with one participant questioning the absence of an upward electric field component from charges on either side of the rod.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement regarding the application of Gauss's Law to infinite versus finite rods, with some asserting the role of symmetry while others remain uncertain about the implications of charges outside the Gaussian surface. The discussion does not reach a consensus.

Contextual Notes

Participants highlight potential limitations in understanding the behavior of electric fields in relation to the geometry of the charged rod and the Gaussian surface, as well as the implications of symmetry in the analysis.

FS98
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To find the electric field from an infinitely long charged rod you can use gauss’s law with a cylinder as your Gaussian surface. I don’t quite understand by this works. Wouldn’t the electric field given by the equation only be the electric field cause by the charge within the cylinder? And if that’s the case, how could gauss’s law describe the charge of an infinite rod with a Gaussian cylinder of finite size?
 
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By symmetry, you know that the field will be purely radial. Therefore the size of the cylinder doesn't matter, since the charge enclosed and the length of the cylinder both scale linearly. You can also reduce the problem by taking a cut perpendicular to the rod, leaving a 2D disc of charge, and use a circle as a Gaussian "surface".
 
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DrClaude said:
By symmetry, you know that the field will be purely radial. Therefore the size of the cylinder doesn't matter, since the charge enclosed and the length of the cylinder both scale linearly. You can also reduce the problem by taking a cut perpendicular to the rod, leaving a 2D disc of charge, and use a circle as a Gaussian "surface".
I still don’t quite understand why the equations would yield different results for a finite and infinite rod. For a finite rod, the charge enclosed would be the same and the area of the Gaussian surface would be the same, so I would think that the electric field would also be the same. But in the case of the infinite rod, there are charges outside of the Gaussian cylinder that would cause a vertical electric field.
 
FS98 said:
But in the case of the infinite rod, there are charges outside of the Gaussian cylinder that would cause a vertical electric field.
The charges on outside the Gaussian cylinder on one side cancel out the field created by the charges on the other side. The field can only be perpendicular to the rod.
 
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DrClaude said:
The charges on outside the Gaussian cylinder on one side cancel out the field created by the charges on the other side. The field can only be perpendicular to the rod.
I understand why the horizontal component would cancel out, but I’m not sure why there still wouldn’t be an electric field in the upward direction from each side of the rod.
 

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