# Gauss’ law and an infinite rod

• I
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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DrClaude
Mentor
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".

FS98
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.

DrClaude
Mentor
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.

FS98
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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