# Calculating Electric Field from a Potential Difference w/o Charge

So here is my homework question:

Two long cylindrical shells of metal (radii $r_1$ and $r_2$, $r_2 > r_1$) are arranged coaxially. The plates are maintained at the potential difference $\Delta\phi$. The region between the shells is filled with a medium of conductivity $g$. Use Ohm's law, $J = gE$, to calculate the electric current between unit lengths of the shells.

So I understand how to go from the current density to the actual current, but I need to find the electric field $E$ in order to find the current density $J$. I am not given the charges on each cylinder. How do I calculate the $E$ field when I am only given a potential difference? I know $\Delta\phi = \int E\,\mathrm{d}\ell$ but I feel like that doesn't help me much.

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What if you postulated some charge $\lambda$ per unit length, found the electric field (using e.g. Gauss' law), calculated the potential difference, and then solved for $\lambda$ in terms of $\Delta\phi$? –  Emilio Pisanty Oct 14 '13 at 0:01
@Emilio I think that would make a good answer (considering that it's a homework question). –  David Z Oct 14 '13 at 0:20
Ah I see what your saying. I'll try that and see what I come up with. I'm still curious about whether or not its actually possible to find the electric field with only the information I'm given or if, like you said, I have to postulate a charge distribution. If I'm not mistaken there physically has to be some sort of distribution, but I wonder whether or not it is necessary information. –  Mr. Frobenius Oct 14 '13 at 0:21
@DavidZ Yes, but I'm on a tablet right now and latex is un.be.lie.va.bly painful. If this doesn't have an answer tomorrow I'll write one up. In the meantime, I suggest Mr Frobenius give my suggestions a try. –  Emilio Pisanty Oct 14 '13 at 0:23
@Emilio No, I meant literally the exact same text you posted as a comment should have been posted as an answer IMO. I'm not saying you should post a full explanation of how such a calculation would work - in fact, an answer like that would be a violation of our homework policy. –  David Z Oct 14 '13 at 0:26