Problem 104 - University Physics with Modern Physics

Question

CP A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P21. 104 ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z-\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\vec{E} .\) Consider points both above and below the annulus in Fig. P21. \(104 .\) (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis. The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section \(14.2 .\) The annulus is held stationary.)

Step-by-Step Solution

Answer

The total electric charge is \(\sigma (\pi R_{2}^2 - \pi R_{1}^2)\). The electric field on the \(x\)-axis is found by integrating contributions from concentric rings. Close to the origin, \(E\) is proportional to \(x\) and the oscillation frequency is \(\frac{1}{2\pi}\sqrt{\frac{q\sigma}{2m\varepsilon_0} (R_2 - R_1)}\).

Step 1: Calculate Charge on the Annulus

The charge on the annulus can be found by calculating the area of the annulus and multiplying it by the surface charge density \(\sigma\). The area \(A\) of the annulus is given by the difference in the areas of the outer and inner circles: \[A = \pi R_{2}^2 - \pi R_{1}^2\]Thus, the total charge \(Q\) on the annulus is:\[Q = \sigma (\pi R_{2}^2 - \pi R_{1}^2)\]

Step 2: Calculate Electric Field on the x-axis

To find the electric field \(\vec{E}\) on the \(x\)-axis, consider the symmetry of the problem and use integration. The field contribution \(d\vec{E}\) from a ring element of radius \(r\) and thickness \(dr\) is:\[dE = \left(\frac{\sigma \cdot 2\pi r \, dr}{4\pi\varepsilon_0}\right) \frac{x}{(x^2 + r^2)^{3/2}}\]Integrate from \(R_1\) to \(R_2\):\[E = \int_{R_1}^{R_2} \frac{\sigma r x}{2\varepsilon_0 (x^2 + r^2)^{3/2}} \, dr\]

Step 3: Evaluate Electric Field Close to the Origin

For points very close to the origin such that \(x \ll R_1, R_2\), we can approximate the expression from Step 2 using a Taylor expansion or small angle approximation,\[E \approx \frac{\sigma x}{2\varepsilon_0} (R_2 - R_1)\]This shows that the electric field is proportional to \(x\), confirming the problem's requirement for points close to the origin.

Step 4: Determine Frequency of Oscillation

Using the approximation from Step 3 where the electric field \(E\) is proportional to \(x\), the force \(F\) on the particle is:\[F = -qE = -\frac{q\sigma x}{2\varepsilon_0} (R_2 - R_1)\]This is equivalent to Hooke's law \(F = -kx\), where \(k = \frac{q\sigma}{2\varepsilon_0} (R_2 - R_1)\). The angular frequency \(\omega\) is given by:\[\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{q\sigma}{2m\varepsilon_0} (R_2 - R_1)}\]The frequency \(f\) is then:\[f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{q\sigma}{2m\varepsilon_0} (R_2 - R_1)}\]

Key Concepts

Surface Charge Density

Surface charge density, denoted by the Greek letter \(\sigma\), is a measure of how much electric charge is distributed over a given area. It is an important concept in electrostatics, as it characterizes the amount of charge on surfaces like the annular disk in this exercise. When defining surface charge density, we use the formula:\[\sigma = \frac{Q}{A}\]Where \(Q\) is the total charge, and \(A\) is the area. For our annular disk, \(A\) is determined by the difference between the outer circle area \(\pi R_2^2\) and the inner circle area \(\pi R_1^2\). Thus, this gives us the total charge as:\[Q = \sigma (\pi R_2^2 - \pi R_1^2)\]

• It is crucial to identify the role of surface charge density in defining electric fields.
• Understanding \(\sigma\) allows us to relate the physical characteristics of the disk to the behavior of electric fields in its vicinity.

Recognizing this relationship is key to solving electrostatic problems involving surfaces like annuli.

Electric Field Integration

Electric field integration is a method used to calculate the electric field by integrating the contributions from infinitesimal charge elements together. It is especially useful in scenarios where the charge distribution isn't concentrated at a point, such as with our annular disk.To find the electric field \(E\) on the x-axis due to the annular disk, we need to integrate the contributions from each infinitesimal within the disk. Using symmetry and integration, the expression for the elemental electric field \(dE\) from a ring element is derived. This involves integrating over the range from \(R_1\) to \(R_2\):\[E = \int_{R_1}^{R_2} \frac{\sigma r x}{2\varepsilon_0 (x^2 + r^2)^{3/2}} \, dr\]

• Electric field integration helps bridge the gap between continuous charge distributions and the resulting fields.
• It shows how different parts of the annulus contribute to the overall electric field at a point along the x-axis.

By understanding this integration process, students can better grasp how complex charge configurations affect electric field patterns.

Oscillation Frequency of Charged Particle

The oscillation frequency of a charged particle in an electric field describes how frequently the particle moves back and forth under the influence of that field. In this problem, we have a negatively charged particle situated close to the center of an annular disk, experiencing forces due to the electric field.For a particle near the center, the electric field is approximately proportional to the distance from the center (\(x\)), allowing us to draw parallels to Hooke's law, which describes simple harmonic motion. This results in:\[F = -qE = -\frac{q\sigma x}{2\varepsilon_0} (R_2 - R_1)\]Here, the negative sign indicates a restoring force, similar to that in a spring system. The resulting angular frequency \(\omega\) and frequency \(f\) of oscillation are:\[\omega = \sqrt{\frac{q\sigma}{2m\varepsilon_0} (R_2 - R_1)}\quad\text{and}\quad f = \frac{\omega}{2\pi}\]

• This equation shows how the physical parameters of the system—mass, charge, and geometry—affect oscillations.
• The notion of resonant frequency and simple harmonic motion in physics can be applied in various engineering and scientific contexts.

This concept not only explains particle motion but also opens the door to exploring oscillatory systems in broader physics applications.

Physics Problem-Solving

Physics problem-solving often involves breaking complex problems into manageable steps, as shown in the annular disk electric field problem. Understanding the core principles and applying systematic approaches enables students to tackle similar exercises effectively.

• Identify knowns and unknowns: Start by clearly stating what information is available and what needs to be determined.
• Apply physical laws: Use principles like Gauss's Law, symmetry, and integration to frame the problem.
• Use approximations wisely: Simplify the problem with approximations for scenarios like small distances, where proportions can simplify equations.

By following these steps consistently, students can find confidence in solving physics problems, reinforcing crucial fundamental concepts needed for more advanced studies. Emphasizing a solid understanding of problem-solving strategies fosters a deeper comprehension of physical phenomena and how theoretical concepts translate into real-world applications.