- #1
Taylor T
- 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