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

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3. Cyosis

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

Last edited: May 18, 2009
4. wgdtelr

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

5. wgdtelr

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

6. 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.

7. gabbagabbahey

5,009
Surely you mean $da=rd\phi dz$....right?

Sure you can, just do the angular integral first

Last edited: May 18, 2009
8. gabbagabbahey

5,009
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!

9. wgdtelr

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

10. 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!

11. gabbagabbahey

5,009
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.