Force on a charge from a cylinder of charge

[kpou]

Homework Statement

Find the force on a charge q a distance r > a away from a cylinder of charge with radius a. The cylinder has a charge density per volume \rho

Homework Equations

E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V} \frac{\rho(r')}{\eta^2}\eta[hat]d\tau'

The Attempt at a Solution

I am not sure how on Earth to start the problem without thinking about the cylinder as infinite length or having q at a midpoint in the cylinder. I don't even know if I am supposed to know how to do that yet :/ What I have done is

\eta=\sqrt{x^2+y^2+z^2}

Does this make
E(r)=\frac{1}{4\pi\epsilon_{0}} \int_V \frac{\rho}{\sqrt{x^2+y^2+z^2}} \frac{xdx+ydy+zdz}{\sqrt{x^2+y^2+z^2}}

Any nudge in the right direction would be great. Thanks :D

Note: formatting gets very easy after you mess up a whole bunch :)

Edit: This problem is near the Dirac delta function section. Perhaps this may help? Unfortunately I don't see how it could fit just for the fact that the dimensions of the cylinder will affect the point charge.

 

[jdwood983]

Is the charge density of the cylinder uniform or is there a function given for it?

 

[kpou]

It is uniform I assume.

 

[jdwood983]

And the problem does not indicate that it is an infinite cylinder?

Also, this problem will likely be loads easier if you work in cylindrical coordinates, rather than Cartesian.

 

[UgOOgU]

E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V} \frac{\rho(r')}{\eta^2}\eta\d\tau'

This equation is not OK. The \eta in the upper part of the fraction do not exists.

 

[kpou]

I'm sorry it is supposed to have a hat over it.

 

[UgOOgU]

No, the equation is really without it.
E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\frac{\rho(r')}{\eta^2} d\tau'

 

[kpou]

And the problem does not state it is infinite, I actually copied it word for word in case I missed something.

 

[jdwood983]

Unless someone suggests otherwise, I'd assume that it is infinite. In this case, you can use Gauss's Law to find the electric field:

<br /> \oint \mathbf{E}\,d\mathbf{a}=\frac{Q_{enc}}{\varepsilon_0}\rightarrow\mathbf{E}=\frac{Q_{enc}}{2\pi\varepsilon_0r}\hat{\mathbf{r}}<br />

 

[kpou]

I am just going by what my book has written for the E equation. It does include \widehat{\eta}.

 

[kpou]

bah, I just noticed the next question specifies an infinitely long solenoid. Maybe it wasn't just forgotten like I was hoping.

 

[UgOOgU]

I am just going by what my book has written for the E equation. It does include \widehat{\eta}.

Ok, now I understand, \widehat{\eta} is the versor of the vector \vec{\eta}.

Sorry, I think that it was wrong because of your substitution in this equation

E(r)=\frac{1}{4\pi\epsilon_{0}} \int_V \frac{\rho}{\sqrt{x^2+y^2+z^2}} \frac{xdx+ydy+zdz}{\sqrt{x^2+y^2+z^2}}

that had to be
\vec{E}(r)=\frac{1}{4\pi\epsilon_{0}}\int_V\frac{\rho}{x^2+y^2+z^2}dx dy dz.