The space between two coaxial cylinders, whose radii are a and b (where `a ltb` as in (figure shown) is filled with a conducting medium . The specific conductivity of the medium is `sigma`.

(a) Compute the resistance along the lenght of cylinder .
(b) Compute the resistance between the cylinders are very long as compared to their radii , i.e ., `Lgtgt` b, where L is the lenght of the cylinders.

Explanation

(a) `R=(rhol)/(A)=(l)/(sigmaA)=(l)/(sigma(pib^(2)-pia^(2)))=(l)/(pisigma(b^(2)-a^(2))`
(b) From Ohm's law , we have
`vec(J)=sigmavec(E)` Assuming radial current density . `vec(J)` becomes
`vec(J)=(I)/(2pirL)hatr" for a"ltrltb`
and , therefore , " " `vec(E)=(I)/(2pisigmarL)hatr`
Here we have used the asssumption that `Lgtgtb` so that `vec(E)andvec(J)` are in cylindrically symmetric form. The potential drop across the medium is thus :
`V_(ab)=-int_(b)^(a)vec(E)(r).vec(dr)=-(I)/(2pisigmaL)int_(b)^(a)(dr)/(r)=(I)/(2pisigmaL)ln ((b)/(a))`
The resistance
`R_(ab)=(V_(ab))/(I)=(In((b)/(a)))/(2pisigmaL)`