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I don't remember enough from my electromagnetism course and I can't find any simple, full example on this subject.

I know, that I can consider the cylinder as a wire with the same charge density (when outside of the cylinder), so I consider the wire case with the same charge density, $\lambda$.

I set the $y$ direction to pass through the wire, and the $x$ direction to pass through the point charge and perpendicular to the wire. Everything is in the $z=0$ plane. Hence, the particle is at $x=r$ position.

From Gauss law, I learn that the field at a point at distance $r$ from the wire I have a field which scales like $\lambda/r$. To find the potential, I need to solve the integral

$V = -\int_b^r E dl = - k \int_b^r \lambda/x dx$

Where $k$ holds the constants and $b$ being a point where the potential is 0.

The solution of this integral is

$V = - k \lambda \log (1/r) + C$

When $b=1$ the potential is 0 so the potential is

$V = k \lambda \log (1/r)$

and the energy is

$U = k q \lambda \log (1/r)$

Here are my questions:

  1. Is any of this true?

  2. I tried to derive this by starting with coulomb law and calculating the energy for a segment of the wire, I get an answer which scales like $1/r$, can you derive the answer for this?

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This looks more or less right to me, and is the approach I'd take. –  Dave Dec 3 '12 at 21:14
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1 Answer

up vote 1 down vote accepted

I think you have a little typo in your formula for $V$. It should read $V=-k \lambda log(r)+C$.

Apart from this if you call $d=\sqrt{r^2+y^2}$ the distance of the point charge in (r,0,0) from a point (0,y,0) on the wire, Coulomb law gives the electrical field in the point (r,0,0) as a vector along the x axes proportional to $$E=2\int_0^{\infty} \frac{\lambda}{d^2} \frac{r}{d}dy=2\int_0^{\infty}\lambda \frac{r}{(r^2+y^2)^{\frac{3}{2}}}dy $$ ($r/d$ is the cos of the angle $\theta$ between the vector pointing from (0,y,0) to (r,0,0) and (r,0,0),and for symmetry you only need to integrate on the upper part of the wire). With the substitution $y=r \ tan(\theta)$, and $dy=r \frac{1}{cos(\theta)^2} d\theta$ this becomes

$$E=2\int_0^{\infty}\lambda \frac{r}{(r^2+y^2)^{\frac{3}{2}}}dy = 2\frac{\lambda}{r}\int_0^{\pi /2}cos(\theta)d \theta= 2\frac{\lambda}{r}$$

so you recover the same electrical field you get from Gauss theorem. The rest is the same..

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