# Total induced charge of an infinite cylindrical conductor

1. ### wgdtelr

9
1. The problem statement, all variables and given/known data
calculate total induced charge on a charged cylinder. where the surface charge density is given by sigma= 2eEo cos(phi)

2. Relevant equations
the total induced charge on the cylinder is

Integral of (sigma) da

can u calculate this integral fo me ... it very urgent..

3. The attempt at a solution

2. ### Cyosis

1,495
Use cylindrical coordinates $da \Rightarrow rd\phi dz$.

Last edited: May 18, 2009
3. ### wgdtelr

9
Can u please give me the limits under which i'Ve to integrate for r ,Phi, z.

4. ### wgdtelr

9
here it is infinite long cylinder and what limits can we take in z- direction.

5. ### Cyosis

1,495
You can't calculate the total charge on an infinite conductor. What you have to do is integrate over a finite piece of conductor then divide the total charge by the length of your z interval. This way you get the total charge on the conductor per length.

6. ### gabbagabbahey

5,013
Surely you mean $da=rd\phi dz$....right?

Sure you can, just do the angular integral first

Last edited: May 18, 2009
7. ### gabbagabbahey

5,013
Well, the entire surface is at some constant radius, so there is no need to integrate over $r$ at all.

If the cylinder is infinitely long, then the limits for $z$ are $\pm \infty$

And the limits for $\phi$ are $0$ to $2\pi$.....These should all be fairly obvious to you....have you not used cylindrical coordinates before?

As for the integration, do the angular integral first!

8. ### wgdtelr

9
yaaaaa I've got zero.. after doin the angular part.

9. ### Cyosis

1,495
Ugh not very handy of me to write down the volume element, thanks.

Actually looking at the function that is to be integrated might help next time!

10. ### gabbagabbahey

5,013
And this result should be no surprise, since an induced charge density does not change the total charge on a surface, it merely redistributes the charges. If the conductor was neutral before the charge was induced, it will still be neutral afterwards.

Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?