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I have two concentric cylinders: A and B, both have a length $L$ (In fact, they are very long, but the problem says that i must only consider this section). I have that $R_b > R_a$ (radius). These two concentric cylinders are separated by a material whose conductivity is $\sigma$, and a difference of potential $V$ is established between them. What is the current that goes from one cylinder to another?

First of all, i did solved this problem, but i did it using current density. My new plan is to use $V = Ri$, and using an integral to calculate the resistance and then use ohm's law. I did it: $ R = \rho \frac{L}{A} $, so $ R = \frac{M}{\sigma A} $, where M is the difference of the radius $b-a$. From this i got to: $R(r) = \frac{M}{\sigma 2 \pi r L} $. Then $$ R = \int_{a}^{b} \frac{M}{\sigma 2 \pi r L} dr $$. and using $V = Ri$, i got to: $$ i = \frac{2 \pi r L \sigma}{M \ln{\frac{b}{a}}} V $$ The answer is exactly like that, however, there is no $M$ in there. My questions are:

I) Can i use $V=Ri$ in this case?

II) The change from $dR$ to $dr$ in the integral would not solve the problem (as i have tried that), so, is there a conceptual mistake in this? I need some help. Thanks!

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I) Can i use $V=Ri$ in this case?

Yes, Ohm's law should apply.

II) The change from dRdR to drdr in the integral would not solve the problem (as i have tried that), so, is there a conceptual mistake in this?

Yes. You are on the right track in computing the resistance using symmetry and integrating the resistance of thin concentric shells of material of radius $r$ over the radius $r$ from $a$ to $b$. However, just by inspecting your formula, notice that it doesn't make any sense for the resistance of a thin shell of material of radius $r$, $R(r)$, to depend on $(b-a)$, because that is completely unrelated to the thin shell itself, so your formula for $R(r)$ is obviously wrong, by a factor of $M$. The rest of it was actually correct.

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  • $\begingroup$ I'm sorry, im not seeing that clearly why does the $M$ is irrelevant to $R$. $\endgroup$ – Vitor C Goergen Sep 30 '16 at 13:39
  • $\begingroup$ $R(r)$ is the resistance of a thin shell of the material between the tubes of radius $r$. Clearly the resistance of this thin shell does not depend on the size of the two cylinders. Where did you get your formula for $R(r)$ from? $\endgroup$ – Suzu Hirose Sep 30 '16 at 13:41
  • $\begingroup$ i doesnt depend on the size of the two cylinders, but i think it depends on how far one cylinder is from another. $\endgroup$ – Vitor C Goergen Sep 30 '16 at 18:49
  • $\begingroup$ I got the formula from $R = \rho \frac{L}{A} $ , when L is the length of a wire. $\endgroup$ – Vitor C Goergen Sep 30 '16 at 18:50
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It is easiest to calculate the resistance $R$ between the inner and outer faces and then use $V=iR$ to find the current.

From $R=\rho \frac{L}{A}$ the resistance of a cylindrical shell of radius $r$, thickness $L=dr$ and height $H$ is $dR=\rho \frac{dr}{2\pi rH}$. The cylinders are in series so the total resistance is the integral
$$ R = \int_{a}^{b} \rho \frac{dr}{2 \pi rH}$$

Your mistake is your formula $R=\frac{b-a}{\sigma A}$. The elementary cylinders used for integration have thickness $L=dr$ not $b-a$. You should insert $b, a$ into the formula after integration, not before.

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  • $\begingroup$ Good explanation. $\endgroup$ – Suzu Hirose Oct 1 '16 at 0:19
  • $\begingroup$ Thank you. I did not understand your answer until after I posted mine. Now I see that our answers are essentially the same! $\endgroup$ – sammy gerbil Oct 1 '16 at 0:21
  • $\begingroup$ You added the things I thought were obvious, like that the shells are in series, so your explanation is clearer than mine. I couldn't understand how OP had got that far towards the solution without seeing the error. $\endgroup$ – Suzu Hirose Oct 1 '16 at 0:23

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