Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I've been working on some exercises and I'm in doubt if my procedure with this one is correct. We have a hollow cylinder with internal radius $r_a$, external radius $r_b$, resistivity $\rho$ and length $L$. A difference of potential $V$ is then applied on the extremes paralel to the axis of the cylinder.

First I'm asked to find the resitance. Well, for this one I've just used that $R=\rho L/A$ and calculated $A=\pi(r_b^2-r_a^2)$. Second, I'm asked to find the current density when $V$ is applied. Well, what I did was suppose that the cylinder is an ohmic resistor so that $V=Ri$ holds, and then I've got $i=V/R$. Then the current density should be $J=i/A$ and so we have $J=V/RA$ which gives $J=V/\rho L$. Is this correct? Can I assume that the cylinder is ohmic?

Third I'm asked to find the electric field. Now I've used that $\rho = E/J$ so that we must have $E = \rho J$ so that $E = V/L$, this says tha that the field is uniform, but I'm a little unsure about it. Finally the fourth one asks to find the resitance again if now the current flows radially from inside to ouside. In this final case I think that it'll change just the cross sectional area, but I didn't find some easy way to write this down.

Can someone give a little help with this?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Your calculations are correct, provided the cylinder is indeed ohmic. The constant $E$ you're getting is the difference in electric field between both terminals.

As for the current flowing from inside to outside, as you said the cross sectional area will be different, and so will the length. The length $L=r_b-r_a$, but the cross sectional area is not uniform, because at the beginning of the wire (the interior), $A=2\pi r_aL$, and at the "end" of the wire (the exterior), $A=2\pi r_bL$. So you'll have to treat each portion of the wire as its own infinitesimal resistor $dR$, and the total resistance is the series combination of them:

$$dR=\rho\frac{dl}{2\pi lL}$$

$$R=\int_{r_a}^{r_b}\rho\frac{1}{2\pi lL}dl$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.