# Total induced charge of an infinite cylindrical conductor

 P: 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
 HW Helper P: 1,495 Use cylindrical coordinates $da \Rightarrow rd\phi dz$.
 P: 9 Can u please give me the limits under which i'Ve to integrate for r ,Phi, z.
 P: 9 Total induced charge of an infinite cylindrical conductor here it is infinite long cylinder and what limits can we take in z- direction.
 HW Helper P: 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.
HW Helper
P: 5,003
 Quote by Cyosis Use cylindrical coordinates $da \Rightarrow rdrd\phi dz$.
Surely you mean $da=rd\phi dz$....right?

 Quote by Cyosis You can't calculate the total charge on an infinite conductor.
Sure you can, just do the angular integral first
HW Helper
P: 5,003
 Quote by wgdtelr Can u please give me the limits under which i'Ve to integrate for r ,Phi, z.
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!
 P: 9 yaaaaa I've got zero.. after doin the angular part.
HW Helper
P: 1,495
 Surely you mean LaTeX Code: da=rd\\phi dz ....right?
Ugh not very handy of me to write down the volume element, thanks.

 Sure you can, just do the angular integral firstp
Actually looking at the function that is to be integrated might help next time!
HW Helper
P: 5,003
 Quote by wgdtelr yaaaaa I've got zero.. after doin the angular part.
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.

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