- #1
Telemachus
- 820
- 30
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 length 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?
Thanks in advance.
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 length 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?
Thanks in advance.
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