A large brass washer has a \(2-\mathrm{cm}\) inside diameter, a \(5-\mathrm{cm}\) outside diameter, and is \(0.5 \mathrm{~cm}\) thick. Its conductivity is \(\sigma=1.5 \times 10^{7} \mathrm{~S} / \mathrm{m} .\) The washer is cut in half along a diameter, and a voltage is applied between the two rectangular faces of one part. The resultant electric field in the interior of the half- washer is \(\mathbf{E}=(0.5 / \rho) \mathbf{a}_{\phi} \mathrm{V} / \mathrm{m}\) in cylindrical coordinates, where the \(z\) axis is the axis of the washer. (a) What potential difference exists between the two rectangular faces? ( \(b\) ) What total current is flowing? ( \(c\) ) What is the resistance between the faces?

For a poloidal electric field \(\mathbf{E}=(0.5 / \rho) \mathbf{a}_{\phi} \mathrm{V} / \mathrm{m}\) in cylindrical coordinates (\(\rho,\phi,z\)), where the \(z\)-axis is the axis of the washer, we can calculate the potential difference \(\Delta V\) as follows: $$ \Delta V = - \int \mathbf{E} \cdot d\boldsymbol{\ell} $$ Here, \(d\boldsymbol{\ell} = \rho d\phi \mathbf{a}_\phi\). Therefore $$ \Delta V = - \int_0^\pi \frac{0.5}{\rho}\rho d\phi = -0.5\int_0^\pi d\phi. $$ Integrating and solving for \(\Delta V\), we get: $$ \Delta V = -0.5[\phi]_0^\pi = -0.5 (\pi - 0) = -\frac{1}{2} \pi \mathrm{V}. $$ So the potential difference between the two rectangular faces is \(\frac{1}{2}\pi \ \mathrm{V}\).

Now we need to find the current density, \(\mathbf{J} = \sigma \mathbf{E}\). Using the provided conductivity \(\sigma = 1.5 \times 10^{7} \mathrm{S/m}\): $$ \mathbf{J} = \sigma \mathbf{E} = \sigma \cdot \frac{0.5}{\rho} \mathbf{a}_{\phi}. $$ To find the total current \(I\), we will integrate the current density over the cross-sectional area of the half-washer. The area of the cross section in the \(\rho-z\) plane is a rectangle with width \(0.5 \ \mathrm{cm}\) and length \((5-2) \ \mathrm{cm} = 3 \ \mathrm{cm}\). Therefore, $$ I = \int_0^{0.5} \int_{0.02}^{0.05} \sigma \cdot \frac{0.5}{\rho} \rho \ d\rho\ dz = 2\sigma \int_0^{0.5} \int_{0.02}^{0.05} d\rho\ dz = 2\sigma \int_0^{0.5} (0.05 - 0.02)\ dz. $$ Solving for the current, we get: $$ I = 2\sigma \times 0.03 \times 0.5 = 1.5 \times 10^{7}\ \mathrm{S/m} \times 0.03\ \mathrm{m} \times 0.5\ \mathrm{m} = 225 \ \mathrm{A}. $$

Finally, we can find the resistance \(R\) using Ohm's law: $$R = \frac{\Delta V}{I}.$$ We've calculated the potential difference \(\Delta V = \frac{1}{2}\pi \ \mathrm{V}\) and the total current \(I = 225 \ \mathrm{A}\). Substituting these values into the equation, we get: $$ R = \frac{\frac{1}{2}\pi}{225} = \frac{\pi}{450} \approx 0.00698 \ \Omega. $$ Thus, the resistance between the two rectangular faces is approximately \(0.00698 \ \Omega\).