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Proving Gauss Law using a "bad" Gaussian surface

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  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data
    What I basically wanna do is to prove Gauss Law with a cylinder perpendicular to an infinite charged wire (I know I can do this simple, but I wanna do it this way)
    EeTd5.png


    This is what I have done so far:

    hm2cR.jpg


    2. Relevant equations
    $$\Phi=\int \frac{dq}{4\pi \varepsilon_0 r^2} \hat{r} \cdot d\vec{A}$$

    3. The attempt at a solution
    So $$d=\frac{z}{tan(\theta)}$$ and therefore $$r^2=z^2 cot^2(\theta)+z^2$$. Now the flux through the top of the cylinder is $$\Phi=\int \frac{\lambda dx}{4\pi \varepsilon_0[z^2(1+cot^2(\theta))]} \widehat{r} \cdot d\vec{A}$$ where
    $$\vec{A}=(\rho d\rho d\phi)\hat{z}$$ and $$\hat{r}=\frac{\vec{r}}{|r|}$$
    So:
    $$\Phi=\int \frac{\lambda dx}{4\pi \varepsilon_0[z^2(1+cot^2(\theta))]^{\frac{3}{2}}} (dcos(\phi),dsin(\phi),z) \cdot (\rho d\rho d\phi)(0,0,\hat{z})$$
    But I don't know how to proceed... Maybe I can use cosine law to find $$\rho$$?
     
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  3. Sep 24, 2016 #2

    Simon Bridge

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    How do you plan to handle the flux through the sides of the cylinder?
    Also: you need a clear statement of what, exactly, it is that you are proving.
     
  4. Sep 24, 2016 #3
    Using the fact that $$d\vec{A}$$ is $$\rho d\rho d\phi \hat{\rho}$$. First, I wanna calculate the flux in the bottom and top of the cylinder, and I expect it will be slightly similar in the case of the sides. What I wanna prove is that this flux is equal to $$\frac{q}{\varepsilon_0}$$
     
  5. Sep 24, 2016 #4

    Simon Bridge

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    OK good luck ... there is a reason people don't try this: it's very difficult.
    It is not uncommon to end up with integrals that cannot be evaluated analytically.

    The integral you are asking about needs limits ...

    ... and you still should make an explicit statement of what you want to prove.
    Are you just rotating the standard gaussian surface for the hell of it or are you trying to find something out?
    ie. why pick a cylinder in that orientation? Why not a spheroid or a cube? Why not have the line of charge pass through the cylinder at an arbitrary angle to the cylinder axis?
     
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