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Electric Flux on a Cylindrical Gauss Surface

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data


    A cylinder of length L and radius R is centered on the z-axis in a region where there is a uniform electric field of E i. Determine the flux for the fourth of the cylindrical surface where x > 0 and y > 0.

    2. Relevant equations

    [tex]\phi = \int E dS[/tex]

    3. The attempt at a solution

    I believe I have the sketch drawn as the problem states.

    If you were to take the entire surface, you'd see:

    [tex]\phi = \int E dS[/tex]
    [tex]\phi = E \int dS[/tex]
    [tex]\phi = E (2\pi RL)[/tex]
    [tex]\phi = 2E\pi RL[/tex]

    I'm confused though. Wouldn't a Guassian surface, such as the cylinder, ultimately have zero flux in the above Electric Field?

    I don't know how to progress using Gauss's Law to cut this into a fourth. I don't see how I can use symmetry to develop a method to use the law.
  2. jcsd
  3. Jul 17, 2009 #2


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    Find the flux through the L by 2*r rectangle that goes through the axis of cylinder and is perpendicular to the E field. The incoming and outgoing flux through the cylinder are both equal to that, right? Sure, that makes zero total flux. But the flux through the two quarter cylinders are both equal.
  4. Jul 18, 2009 #3
    So you're saying I should take a Gaussian rectangle and do this to it:

    [tex]\phi = \int \vec{E} \cdot d\vec{S}[/tex]
    [tex]\phi = E A cos \theta[/tex]
    [tex]\phi = E (2RL)(1)[/tex]
    [tex]\phi = 2ERL[/tex]

    This gives us the flux through the rectangle as 2ERL. The correct answer to the problem is ERL (1/4th of the cylinder). However, I do not follow the reasoning why I took this Gaussian surface or why it's useful in finding the cylinder's flux.
  5. Jul 18, 2009 #4


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    Think about it. All of the E flux coming from the negative x direction (referring to your diagram) first passes through the negative x part of the cylinder, then the rectangle, then the positive part of the cylinder. The volume consisting of the negative x part of the cylinder plus the rectangle has zero total flux. Same for the positive. So flux through the rectangle is the same as the flux through the cylinder halves.
  6. Jul 18, 2009 #5
    I'm a "why" learner. I've always been. I don't understand what I'm doing until I understand why I'm doing them. I don't understand the point of finding these Gaussian surfaces and what Flux is going to get me. Therefore, I have absolutely no inclination on how to approach these problems since I don't know why flux surfaces matter, and as a result, I don't understand how to take the surfaces either.

    I understand your reasoning, but there is absolutely no way that I can come to that conclusion myself. I'm extremely frustrated because I feel that it's over my head, but I know deep down it's not. It's the method by which these classes are taught. I have a really bad feeling about not being able to get through my E&M class in the Fall (my first semester back in college after 10 years, having dropped out for health reasons).

    As for the problem above, if the flux through the rectangle (2R x L) is equal to one half of the cylinder, why can I assume that the flux through 1/2 of the rectangle is equal to 1/2 the flux through the cylinder (the quarter piece)?
  7. Jul 18, 2009 #6


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    Symmetry. The two quarters of the cylinder have the same flux through them. So each must have 1/2 the flux through the half cylinder. Just keep trying to do these problems and try not to get frustrated. That will only make it worse. Once you've convinced yourself you'll never understand, you'll have a hard time unconvincing yourself.
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