# Electrostatics, infinite charged hollowed cylinder

1. Sep 20, 2011

### Telemachus

Hi. I want to know if I did this on the right way. The exercise says: An infinite hollowed cylinder, with the cavity being another concentric cylinder has a uniform charge density. Find the electric field and the potential over all space.

And this is how I proceeded. I've called the inner radius for the cylinder a, and the outer b. Then:
$$\rho=cte=\sigma$$
$$Q=\int _V \rho d\tau=\sigma(b^2-a^2)\pi l$$
Being l the lenght of the cylinder (infinite for the case, this is one of the things that bothers me). Then I've applied Gauss law to get the electric field for the region outside the cylinder, with the radius bigger than b.

$$\oint E da=\frac{Q}{\epsilon_0}$$
$$E2\pi r l=\frac{\sigma(b^2-a^2)\pi l}{\epsilon_0}$$
$$E=\frac{\sigma(b^2-a^2)\hat r}{2\epsilon_0 r}$$

Then the potential for the same region, I choose the origin at the point a. I can't choose infinity, because it isn't a localized charge, and at zero it blowed up too, so I just choose a:
$$V(r)=-\int_o^r Edl=-\int_a^r\frac{\sigma(b^2-a^2)\hat r}{2\epsilon_0 r}=-\frac{\sigma(b^2-a^2)}{2\epsilon_0}\ln \left ( \frac{r}{a} \right )$$

I proceeded similarly for the other regions. The answer I get looks some kind of weird, the natural logarithm doesn't looks good, and it disconcerted me that I couldn't choose the center of the cylinder as the origin, so I think I'm probably making some mistake here. Besides the potential seems to increase when one gets further away from the cylinder, which makes no sense at all.

What you say?

Last edited: Sep 20, 2011
2. Sep 20, 2011

### pabloenigma

Your calculation is correct.It makes no sense at all,because even infinite cylinders dont make any sense.In words of Griffiths,"Notice that the difficulty occurs only in textbook problems; in "real life" there is no such thing as a charge distribution that goes on forever, and we can always use infinity as our reference point."

You are correct,no need to lose sleep over it.

3. Sep 20, 2011

### Telemachus

Thank you Pablo. I just read that words from griffiths like an hour ago :P curious considence. But anyway, it doesn't make any sense I think the expression I get for the potential, the logarithm increases to infinity as the radius increases. Compared to an infinite plane... oh, wait, I was thinking on the field of the plane, now I think that it makes sense, because the logarithm increases slower than the expression that I could get for an infinite plane.

Thanks.

4. Sep 20, 2011

### ehild

Your derivation is correct. You cannot choose the zero of potential at infinity in case of cylindrical charge distribution. Think: Infinite long cylinder does not exist. You can consider it infinite and use symmetry to calculate the electric field if the distance from the axis of the cylinder is much smaller than the length of the cylinder. If the cylinder is finite and you want to determine the electric field and potential at a point, farther than the length of the cylinder, you can not ignore the ends. And very far away the cylinder will behave more an more like a point charge.

ehild