Calculating electric field on a long insulating cylinder?

[coffeem]

[PLAIN]http://img20.imageshack.us/img20/6278/electricfieldquestion.png

[PLAIN]http://img828.imageshack.us/img828/659/electricfieldanswer.jpg

I know I have to use Gauss's law... However I am unsure about how I go about this... I've tried doing this on about 3 sheets of paper and have done nothing but got big R and small r the wrong way around... On top of that I am using values for dA which I am unsure about...

Any help would be appreciated.

 

[Mindscrape]

Well, to be fair to you, this type of problem is usually reserved for upper level undergrad, when everybody has done vector calculus.

You start off with Gauss's law
\int \int \mathbf{E} \cdot d\mathbf{a} = \int \int \int \rho dV / \epsilon_0

So, you can really pick any surface to integrate over, but it might not be the best idea to make a box. Since the electric field will be normal to the cylinder it originates from, it's a good idea to pick a surface that's area vectors will be parallel to the electric field, which gives a... you guessed it, a cylindrical surface. Now you have your Gaussian surface, and you just have to do the integrals in the right coordinate system. For the da vector you will want to piece together the cylindrical line elements that integrate over the surface.

I'm sure you'll struggle a bit more even after I give you this advice, but that's part of the process, it's how you learn best. After you've used up a few more sheets of paper, come back and either I or someone else will give you a hand.

 

[coffeem]

Oh... I was using the spherical symetry to really really really simplify the problem...

Umm... I will post my working so you can tell me where I am doing wrong! thanks

 

[coffeem]

[PLAIN]http://img829.imageshack.us/img829/9453/wrongworking.jpg

 

[Mindscrape]

You have to use a cylindrical surface. Well, you don't have to, but I strongly advise you do so that the dot product of E and A will not have any cosø terms. E*AreaCylinder (Gaussian surface) = triple integral of charge density (throughout the physical cylinder)

Do you know integration?

 

[coffeem]

You have to use a cylindrical surface. Well, you don't have to, but I strongly advise you do so that the dot product of E and A will not have any cosø terms. E*AreaCylinder (Gaussian surface) = triple integral of charge density (throughout the physical cylinder)

Do you know integration?

Yes I know how to integrate functions - just not in this case!

 

[Mindscrape]

Sorry I didn't get back to you, I had to do some traveling this weekend. So anyway, what you will want to do is

E(2\pi r'|_{r'=?} L) = \frac{1}{\epsilon_0} \int_0^L \int_0^{2\pi} \int_{r'=0}^{r'=?} \rho r' dr d\phi dz

determine the right values to evaluate and integrate for r', depending on your 2 different cases.