Question

A thin, flat washer is a disk with an outer diameter of \(10.0 \mathrm{~cm}\) and a hole in the center with a diameter of \(4.00 \mathrm{~cm} .\) The washer has a uniform charge distribution and a total charge of \(7.00 \mathrm{nC}\). What is the electric field on the axis of the washer at a distance of \(30.0 \mathrm{~cm}\) from the center of the washer?

Short answer

Based on the step-by-step solution, create the following short answer: To find the electric field on the axis of a washer with total charge (7.00 nC) at a distance of 30.0 cm from its center, we first determined the charge per unit area using the washer's total charge and area. Then, we divided the washer into infinitesimal rings and found the electric field contributed by each ring. By integrating the electric field contributions from all infinitesimal rings, we calculated the total electric field. Substituting the given values for charge per unit area, distance along the axis, and inner and outer radii, we found that the electric field on the axis of the washer is approximately 8.43 x 10^6 N/C in the positive x direction.

Step-by-step solution

Determine the charge per unit area of the washer

Since the washer has a uniform charge distribution, we can use the total charge of the washer and its area to calculate the charge per unit area. The area of the washer is the difference between the areas of the outer circle and the inner circle, $$ A = A_\text{outer} - A_\text{inner} = \pi R_\text{outer}^2 - \pi R_\text{inner}^2, $$ where \(R_\text{outer}\) is the radius of the outer circle (5.00 cm) and \(R_\text{inner}\) is the radius of the inner circle (2.00 cm). The charge per unit area, \(σ\) is therefore given by $$ σ =\frac{Q_\text{total}}{A}, $$ where \(Q_\text{total}\) is the total charge on the washer.

Divide the washer into infinitesimal charge elements

We will now divide the washer into infinitesimal rings of radius \(r\), each having a width \(dr\). We can then find the electric field contributed by each ring and integrate over the whole washer to get the total electric field. Each infinitesimal ring has an area \(dA = 2\pi r dr\), and its infinitesimal charge, \(dq\), can be determined using the charge per unit area we calculated in Step 1: $$ dq = σ dA. $$

Determine the electric field contributed by each infinitesimal ring

Now, let's find the electric field contributed by each infinitesimal ring. Due to symmetry, each infinitesimal ring will produce an electric field in the same direction at the point where we want to find the electric field (\(30.0 \mathrm{~cm}\) from the center on the axis). The electric field, \(d\vec{E}\) produced by the infinitesimal ring at location x along the axis of symmetry is given by $$ d\vec{E} = \frac{1}{4\pi ε_0} \frac{dq (xd\hat{x})}{(r^2 + x^2)^{3/2}}, $$ where \(ε_0\) is the vacuum electric permittivity and x is the distance (\(30.0 \mathrm{~cm}\)) along the axis.

Integrate to find the total electric field

Our goal is to find the total electric field produced by the washer. We now have an expression for the infinitesimal electric field contributed by each infinitesimal ring. To find the total electric field, we need to integrate the expression with respect to \(r\) over the entire surface of the washer: $$ \vec{E} = \int_{R_\text{inner}}^{R_\text{outer}}d\vec{E}= \frac{1}{4\pi ε_0} \int_{R_\text{inner}}^{R_\text{outer}} \frac{σ 2\pi r dr (xd\hat{x})}{(r^2 + x^2)^{3/2}}. $$ Simplify and solve the integral: $$ \vec{E} = \frac{σx}{2 ε_0} \left[\frac{1}{\sqrt{R_\text{inner}^2 + x^2}} - \frac{1}{\sqrt{R_\text{outer}^2 + x^2}}\right]\hat{x}. $$

Substitute the given values and calculate the electric field

Finally, substitute the given values for the charge per unit area, the distance along the axis, and the inner and outer radii: $$ \vec{E} = \frac{(7.00 \times 10^{-9} \mathrm{C/m^2})(0.30 \mathrm{m})}{2(8.85 \times 10^{-12} \mathrm{C^2/Nm^2})} \left[\frac{1}{\sqrt{(0.02 \mathrm{m})^2 + (0.30 \mathrm{m})^2}} - \frac{1}{\sqrt{(0.05 \mathrm{m})^2 + (0.30 \mathrm{m})^2}}\right]\hat{x}. $$ After calculating, we get: $$ \vec{E} \approx 8.43 \times 10^6\text{ N/C}\hat{x}. $$ Thus, the electric field on the axis of the washer at a distance of \(30.0\mathrm{~cm}\) from its center is approximately \(8.43 \times 10^6\text{ N/C}\) in the positive x direction.

Key concepts

Charge Distribution

Understanding charge distribution is crucial when analyzing electric fields in electrostatics. It refers to how electric charge is arranged or spread out over a physical object. As in the provided example, we worked with a washer that has a uniform charge distribution, meaning the charge is spread evenly across its surface. For uniform charge distributions over two-dimensional shapes, the electric charge per unit area, known as surface charge density (denoted as \(\sigma\)), is a key factor.

By dividing the washer's total charge by its area—after subtracting the hole's area from the total—we calculated this surface charge density. This is a foundational step which guides us towards understanding how much charge is present on different segments of the washer, further allowing us to calculate the electric field it produces at various points in space.

Gauss's Law

Gauss's Law is a powerful principle that relates the distribution of electric charge to the resulting electric field. It states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\epsilon_0\)).

In the context of our washer problem, Gauss's Law isn't directly applied because we're dealing with a non-spherical charge distribution; however, the same underlying principles help us understand how charges give rise to electric fields. In systems with high symmetry, such as spheres, cylinders, or planes, Gauss's Law simplifies the calculation significantly. Even though it wasn't the principal tool in this exercise, it's a fundamental law that students must grasp to succeed in more complex electrostatic scenarios.

Electric Field Intensity

Electric field intensity, or simply electric field, represents the force per unit charge experienced by a tiny positive test charge placed in the field. It's a vector quantity, indicating both how strong the field is and the direction it points. In the example of the washer, we broke down the washer into infinitesimal rings to calculate the electric field each small part contributes at a specific point, which is 30.0 cm away along the axis.

To determine the electric field due to each ring, we needed to integrate these contributions over the entire surface of the washer. This process resulted in a final expression that represents the intensity and direction of the electric field produced by the charged washer at the specified point. This illustrates how electric field calculations can vary depending on geometry and require integral calculus when dealing with continuous charge distributions.

Electrostatics

Electrostatics is the study of electric charges at rest. It encompasses the forces between charges, the fields they generate, and their behavior in materials. Our washer problem is a classic example of electrostatics in action. We were tasked to calculate the electric field at a fixed point due to a static charge distribution.

In this scenario, understanding electrostatic principles, such as the nature of electric fields and how they arise from charges, is vital. The method we used—breaking the problem into smaller, more manageable pieces and then using calculus to find a comprehensive solution—is a common strategy in electrostatics. Problems like these help students learn to apply general electrostatic concepts to specific situations, thereby deepening their understanding of how physics describes the natural world.