Question

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?

Short answer

Question: Calculate the resistance between the two rectangular faces of a half-washer with given parameters. Answer: The resistance between the two rectangular faces is approximately \(0.00698 \ \Omega\).

Step-by-step solution

Potential Difference Calculation

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 \(\frac{1}{2}\pi \ \mathrm{V}\).

Total Current Calculation

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: $$ I = 2\sigma \times 0.03 \times 0.5 = 1.5 \times 10^{7}\ \mathrm{S/m} \times 0.03\ \mathrm{m} \times 0.5\ \mathrm{m} = 225 \ \mathrm{A}. $$

Resistance Calculation

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 \(0.00698 \ \Omega\).

Key concepts

Electric Field

An electric field represents the force per unit charge at a point in space, directing the way in which a positive charge would move. In simpler terms, it describes how charged objects interact with each other through space, even when they are not in direct contact. Electric fields are typically denoted by the vector \(\mathbf{E}\), and measured in volts per meter (V/m).

In the given exercise, the electric field is expressed uniquely in cylindrical coordinates as \(\mathbf{E} = \frac{0.5}{\rho} \, \mathbf{a}_\phi\, \mathrm{V/m}\). This signifies that the electric field strength diminishes inversely with the radial distance (\(\rho\)) from the axis and acts along the circumferential direction (\(\mathbf{a}_\phi\)).

The magnitude indicates how strong the force would be on a positive charge placed at different positions within the field. Understanding the behavior and direction of the electric field is crucial for calculations involving potential difference and current, as it helps predict how charges will move or be affected in the vicinity of this field.

Cylindrical Coordinates

Cylindrical coordinates provide a way to describe points in space, particularly useful when dealing with problems possessing cylindrical symmetry, such as the brass washer in this problem.

In cylindrical coordinates, a point is defined by three values: radius \(\rho\), angle \(\phi\), and height \(z\):

• \(\rho\) - the radial distance from the axis, essentially the distance from the central z-axis to the point.
• \(\phi\) - the angular coordinate, representing the angle around the z-axis, similar to longitude on Earth.
• \(z\) - the height above or below a reference plane, equivalent to latitude on Earth.

The exercise involves integration in these coordinates, acknowledging their importance for expressing fields and performing calculations in systems where a radial symmetry is inherent. Especially in electrical and electromagnetic contexts, these coordinates often simplify the math by aligning with the symmetry of the system being analyzed.

Current Density

Current density provides a measure of the electric current per unit area flowing through a material. It is a vector quantity, represented as \(\mathbf{J}\), and provides insight into how charges are moving within a conductor.

The magnitude of the current density depends on both the electric field present and the material's ability to conduct electricity, defined by its conductivity \(\sigma\).

• Formula: \(\mathbf{J} = \sigma \mathbf{E}\), where \(\mathbf{E}\) is the electric field and \(\sigma\) is the conductivity.
• In the exercise: Given \(\sigma = 1.5 \times 10^7 \, \mathrm{S/m}\), indicating a good conductive material.

To find the total current in the washer, one must calculate \(\mathbf{J}\) over the cross-sectional area and integrate. This shows how much charge moves per second across a given cross-section, ultimately leading to the current \(I\), showing how quickly electrons are transported through the washer when a voltage is applied.

Ohm's Law

Ohm's Law is a foundational principle in electromagnetism that relates the voltage across a conductor, the current flowing through it, and the resistance it presents. It is usually expressed as \(V = I \cdot R\) where \(V\) is the voltage, \(I\) the current, and \(R\) the resistance.

This law simplifies the analysis of practical electrical systems by providing a straightforward way to calculate one variable if the other two are known. Resistance, in particular, quantifies how much a material opposes the flow of electric current.

In the context of the exercise:

• Voltage difference obtained as \(\Delta V = \frac{1}{2}\pi \, \mathrm{V}\)
• Total current calculated as \(I = 225 \, \mathrm{A}\)
• Resistance formula used: \(R = \frac{\Delta V}{I}\)

The values computed offer a real-world application of Ohm's Law, with resistance being critical for determining how efficiently electric devices like the washer can transmit currents amid applied voltages. Understanding these interactions helps to design systems that use energy more efficiently and avoid electrical overloads.