# Total induced charge of an infinite cylindrical conductor

1. May 18, 2009

### wgdtelr

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. May 18, 2009

### Cyosis

Use cylindrical coordinates $da \Rightarrow rd\phi dz$.

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

### wgdtelr

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

4. May 18, 2009

### wgdtelr

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

5. May 18, 2009

### Cyosis

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. May 18, 2009

### gabbagabbahey

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

Sure you can, just do the angular integral first

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

### gabbagabbahey

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. May 18, 2009

### wgdtelr

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

9. May 18, 2009

### Cyosis

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. May 18, 2009

### gabbagabbahey

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.