Question 705060

SCQHARD

A two-dimensional sheet with a uniform sheet conductivity of $\sigma$ has a central metallic point contact and a circular metal contact at the boundary as shown in the figure. If a constant current $I$ is injected through the central contact and collected at the boundary, then the voltage difference between two points on the sheet at radius $r_1$ and $r_2$ is proportional to

(A)

$\frac{I}{\sigma}[\tan^{-1}(\frac{r_2}{r_1}) - \frac{\pi}{4}]$

(B)

$\frac{I}{\sigma}[\ln(\frac{r_2}{r_1})]$

(C)

$\frac{I}{\sigma}(\frac{r_2-r_1}{r_2+r_1})$

(D)

$\frac{I}{\sigma}(\frac{r_2-r_1}{r_2+r_1})^3$

Detailed Solution

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 $\ln(r_2/r_1)$ .

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