Electric Field from Long Hollow Cylinder

[FS98]

Homework Statement

Consider the hollow cylinder from Exercise 1.59. Use Gauss’s law to show that the field inside the pipe is zero. Also show that the field outside is the same as if the charge were all on the axis. Is either statement true for a pipe of square cross section on which the charge is distributed with uniform surface density?

The cylinder is long.

Homework Equations

The Attempt at a Solution

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I have no idea how to do this problem. Can somebody walk me through it? I’m assuming I have to use the second equation here, but I’m not quite sure what some of it means or how to apply it.

 

[gneill]

If you take a moment to reflect on what exactly Gauss' Law says, and in particular what Q is defined to be, then you should be able to make a good attempt at solution, if not solve the problem entirely.

 

[FS98]

If you take a moment to reflect on what exactly Gauss' Law says, and in particular what Q is defined to be, then you should be able to make a good attempt at solution, if not solve the problem entirely.

Can I get phi by multiplying the volume of the cylinder by ρ then dividing by e0 to get pi(r^2)(h)(ρ)/e0?

Then do I plug that into the other formula? If so how? I don’t know what the double integral is and I’m unsure what to do with the da. Is that a dot product or is the da just letting you know what to integrate with respect to?

 

[gneill]

The double integral represents a surface integral, that is, over the Gaussian Surface. As such it's an integration over an area, hence the dA represents a differential "patch" of that surface area.

Take a look at the following PDF to review: http://bolvan.ph.utexas.edu/~vadim/classes/15s/GaussLaw.pdf
In one section it considers a hollow cylinder.

You need to choose suitable Gaussian surfaces (presumably taking advantage of the symmetry inherent in the scenario) that will allow you to evaluate the integral easily (essentially by inspection).