# Proving Gauss Law using a "bad" Gaussian surface

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1. Sep 24, 2016

### math4everyone

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)

This is what I have done so far:

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 $$\rho$$?

2. Sep 24, 2016

### Simon Bridge

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.

3. Sep 24, 2016

### math4everyone

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 $$\frac{q}{\varepsilon_0}$$

4. Sep 24, 2016

### Simon Bridge

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.