- #1

- 6

- 0

## Homework Statement

WITHOUT using Gauss's law calculate the electric field at all points outside and inside an infinite cylinder with circular cross section of radius b and charge density ρ

Hint: Use cylindrical coordinates and place the target charge on the x axis..

## Homework Equations

[tex]

E = \frac{1}{4\pi \varepsilon _{0}}\int _{v}\frac{\rho\vec{R} }{\left \| R \right \|^{3}}

[/tex]

## The Attempt at a Solution

I started with trying to find the field outside first:[/B]

I figured the best way to do this would be to find R between any point inside the cylinder to the point outside in the following manner. Let r(r,0,0) lie on the x axis.

[tex]

\vec{r}=r\hat{r}+0\hat{\theta}+0\hat{z} \\

\vec{r'}=r'\hat{r}+\theta\hat{\theta}+z\hat{z}\\

[/tex]

Let P be the vector between r'\hat{r} and r using the law of cosines:

[tex]

P = \sqrt{r'^{2}+r^{2}-2rr'cos(\theta)}

[/tex]

Let h be the z height from r' to r

[tex]h = (0-z')\hat{z}[/tex]

thus R by the pythagorean theorem must be [tex] R = \sqrt{h^{2}+P^{2}}[/tex]

simplifying and plugging in R

[tex] R = \sqrt{(-z)^{2}+r'^{2}+r^{2}-2rr'cos(\theta)}[/tex]

at this point I am a little bit lost, I know that the R value should be a vector but it is a scalar, I know that the field will be only in the [itex]\hat{r}[/itex] direction. the differential charge [tex]dq = \rho dv=\rho r'd\theta'dr'dz'[/tex]

I'm not sure where I am going wrong, The integral I would end up with is:

[tex]

E = \frac{1}{4\pi \varepsilon _{0}}\int_{z=-\infty }^{\infty}\int_{r=0}^{a}\int_{\theta=0}^{2\pi}\frac{\rho R}{R^3}\hat{r}

[/tex]

This integral is not only difficult but goes to zero as the z^2 integrated at both inifinites will send the integral to zero.

Here is an image I have drawn for convenience: https://i.sli.mg/CRVdar.png

Thank you in advance for your help