Question 705060

For a 2D sheet with radial symmetry, the current density $J = \frac{I}{2\pi r \delta}$. Using Ohm's law $J = \sigma E$, we have $E = \frac{I}{2\pi r \sigma \delta}$. The potential difference $V = \int_{r_1}^{r_2} E dr = \int_{r_1}^{r_2} \frac{I}{2\pi r \sigma \delta} dr = \frac{I}{2\pi \sigma \delta} \ln(\frac{r_2}{r_1})$.

The term $\delta$ is absorbed into the effective conductivity for a sheet, thus the result is proportional to $\ln(r_2/r_1)$.