# Electric potential on a uniformly charged tube

A hollow cylinder, with no top or bottom, of radius $R$ and length $L$ is uniformly charged with density $\sigma>0$. I have to find the point on space where a point charge $q>0$ has to be drawn from $\infty$ with speed $v\rightarrow0$ so that the work done is maximum, so I have to find the point with maximum potential. What I did was find the electric field, using Gauss's law, and then set $V(\infty)=0$:

$0<\rho < R: E=0 \\R<\rho < \infty: E=\frac{\sigma R}\rho$

so $V(\rho)=-\int\limits_\infty^\rho{E.dl}$ .

For $R<\rho<\infty$ I get $V(\rho)=\sigma R \ln(\infty/\rho)$, so here I don't know what to do. Maybe I shouldn't have used Gauss's law, and calculate $E$ the 'classic' way, integrating over the cylinder, but the integral looks so complicated. This problem was given on an exam and trust me, if the integral is actually that complicated, I wasn't supposed to do it that way. So, what should I do?

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Your electric field is wrong. The cylinder is not infinitely long and so there is no translational invariance along the $z$-direction. The electric field will have components along the $z$-direction as well. –  suresh Feb 13 '14 at 2:45
How would the integral for calculating $E$ be, then? Could you please post it, not solved if you don't want, but to see if what I'm doing is OK. (Even Mathematica/WolframAlpha doesn't respond when asked for this integral) –  user3272994 Feb 13 '14 at 3:32
Try to do something simpler. Compute the electric field along the axis of the cylinder -- this is easy to do. The integral that you need to do to compute the potential at the point $(x,0,z)$ is up to multiplicative factors $\phi(x,0,z)\propto \int_0^{2\pi} d\varphi \int_{-L/2}^{L/2}dw\ (x^2+R^2-2xR \cos\varphi +(z-w)^2)^{-1/2}$ where I have set $y=0$ using cylindrical symmetry in the problem. You can replace $x$ by $\rho$ at the end. –  suresh Feb 13 '14 at 4:32
But how would I show that the required point is on the axis of the cylinder? –  user3272994 Feb 13 '14 at 11:39