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Oscillating in a ring A ring with radius \(R\) has uniform positive charge density \(\lambda\). A particle with positive charge \(q\) and mass \(m\) is initially located at the center of the ring and is then given a tiny kick. If it is constrained to move in the plane of the ring, show that it undergoes simple harmonic motion (for small oscillations), and find the frequency. Hint: Find the potential energy of the particle when it is at a (small) radius, \(r\), by integrating over the ring, and then take the negative derivative to find the force. You will need to use the law of cosines and also the Taylor series \(1 / \sqrt{1+\epsilon} \approx 1-\epsilon / 2+3 \epsilon^{2} / 8\)

Short Answer

Expert verified
Yes, the particle undergoes simple harmonic motion and the frequency of its oscillation is given by \(f = \frac{1}{2 \pi} \sqrt{\frac{q \lambda R }{4 \pi \epsilon_0 m}} \)

Step by step solution

01

Find the Potential Energy

The potential energy (U) of the particle, when it is at a small radius, r, can be obtained by integrating over the ring. As such, the potential energy due to a small charge dq in the ring at an angle theta is: \(dU= \frac{1}{4 \pi \epsilon_{0}} \frac{q dq}{\sqrt{R^{2}+r^{2}-2Rr \cos \theta}}\). Here, dq = \(\lambda Rd\theta\) since dq is a small section of the ring with charge density \(\lambda\). Substituting for dq and integrating over the ring (0 to 2pi), we get: \(U= \frac{1}{4 \pi \epsilon_0} \frac{q \lambda R}{\sqrt{R^{2}-r^{2}}}\)
02

Derive the Force

The force experienced by the charged particle can be obtained by taking the negative derivative of the potential energy with respect to r: \(F= -\frac{dU}{dr}\). Substituting the value of U from Step 1 and differentiating, we get the expression for the force: \(F= \frac{1}{4 \pi \epsilon_0} q \lambda R^{2} \frac{r}{(R^{2}-r^{2})^{3/2}} \).
03

Set up the differential equation for SHM

For a system in SHM, the net force acting on the system should be proportional to the displacement and should be in the opposite direction. Therefore, we equate the derived force to mass times acceleration (which is proportional to r for SHM) to get the differential equation: \(m \frac{d^{2}r}{dt^{2}}=- \frac{1}{4 \pi \epsilon_{0}} q \lambda R^{2} \frac{r}{(R^{2}-r^{2})^{3/2}} \)
04

Solve the differential equation for small oscillations

The problem hints at taking advantage of the Taylor series expansion of \(\frac{1}{\sqrt{1+\epsilon}} \approx 1- \epsilon / 2 + 3 \epsilon^{2} / 8\) for small oscillations, where \( \epsilon = \frac{r^{2}}{R^{2}}\) . By substituting this expansion in our differential equation, and keeping only terms up to first order in \( \epsilon \) (since we are considering small oscillations), we get an equation of motion similar to that of SHM: \( \frac{d^{2}r}{dt^{2}}= - \frac{q \lambda R^{3}}{4 \pi \epsilon_0 m R^{2}} r \)
05

Find the Frequency of Oscillation

One can easily recognize the equation above as similar to that of Simple Harmonic Motion, i.e., \(\frac{d^{2}x}{dt^{2}} = -\omega^{2}x \) where \(\omega\) is the angular frequency. Comparing the two equations, the frequency \(f\)=\(\frac{\omega}{2\pi}\) of the particle can be expressed as \(f = \frac{1}{2 \pi} \sqrt{\frac{q \lambda R }{4 \pi \epsilon_0 m}} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
When talking about potential energy in the context of charged particles, it's all about the energy stored due to their position in an electric field. In this scenario, the charged particle is at a small radius, \( r \), from the center of the charged ring. To find the potential energy \( U \), we integrate the effect of tiny charges \( dq \) in the ring over the entire 360 degrees.

The expression used here, \( dU = \frac{1}{4 \pi \epsilon_{0}} \frac{q \, dq}{\sqrt{R^{2}+r^{2}-2Rr \cos \theta}} \), considers the interaction between a tiny piece of the ring and the particle.

Remember, the potential energy increases with proximity to the source of the electric field. By integrating, we obtain the total energy from all sections of the ring, expressed as \( U= \frac{1}{4 \pi \epsilon_0} \frac{q \lambda R}{\sqrt{R^{2}-r^{2}}} \). This formula captures how energy relates to the particle's position, key for analyzing its motion.
Electric Force
Force in physics is about interaction, and when it comes to electric forces, it's all about the charged particles pushing or pulling on each other. The electric force \( F \) on the particle is derived from the potential energy. By taking the negative derivative of the potential energy with respect to \( r \), we determine how the force changes as the position changes.

Here's how it works: \( F= -\frac{dU}{dr} \). This step gives us the force using \( F= \frac{1}{4 \pi \epsilon_0} q \lambda R^{2} \frac{r}{(R^{2}-r^{2})^{3/2}} \).

This formula helps describe how the force acts when the particle moves slightly away from the center. Newton's third law tells us every action has an equal and opposite reaction, helping define this interaction. Understanding this force is crucial as it leads us to discuss oscillatory motion.
Charge Density
Charge density is the distribution of electric charge per unit length, area, or volume. Here, since we're dealing with a charged ring, the charge density, denoted as \( \lambda \), is linear.

Linear charge density means the amount of charge per unit length of the ring. It's vital in these calculations because it helps determine the strength of the electric field created by the ring.

Charge density affects potential energy as it impacts how much influence each segment of the ring has over the charged particle. This concept helps bridge the physical understanding of how closely packed charges interact in space, affecting overall energy and force calculations.
Oscillation Frequency
The frequency of oscillation measures how many complete cycles occur per unit time. In our scenario, the charged particle undergoes simple harmonic motion (SHM) near the center of the charged ring.

From the expression for SHM, \( \frac{d^{2}r}{dt^{2}} = - \omega^{2}r \), we identify \( \omega \) as the angular frequency. It's linked to frequency \( f \) by \( f = \frac{\omega}{2\pi} \).

For small oscillations, the frequency of this motion is given by \( f = \frac{1}{2 \pi} \sqrt{\frac{q \lambda R}{4 \pi \epsilon_0 m}} \). It's all about how quickly the particle oscillates back and forth in this electric field.

Frequency determines the oscillation speed, which boils down to mass, charge density, and the shape of the potential energy landscape. It's essential for understanding how systems behave under small disturbances, linking kinetic and potential energy in a harmonious dance.

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Most popular questions from this chapter

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

Hydrogen atom \(* *\) The neutral hydrogen atom in its normal state behaves, in some respects, like an electric charge distribution that consists of a point charge of magnitude \(e\) surrounded by a distribution of negative charge whose density is given by \(\rho(r)=-C e^{-2 r / a_{0}} .\) Here \(a_{0}\) is the Bohr radius, \(0.53 \cdot 10^{-10} \mathrm{~m}\), and \(C\) is a constant with the value required to make the total amount of negative charge exactly \(e\). What is the net electric charge inside a sphere of radius \(a_{0} ?\) What is the electric field strength at this distance from the nucleus?

Potential energy of a cylinder A cylindrical volume of radius \(a\) is filled with charge of uniform density \(\rho\). We want to know the potential energy per unit length of this cylinder of charge, that is, the work done per unit length in assembling it. Calculate this by building up the cylinder layer by layer, making use of the fact that the field outside a cylindrical distribution of charge is the same as if all the charge were located on the axis. You will find that the energy per unit length is infinite if the charges are brought in from infinity, so instead assume that they are initially distributed uniformly over a hollow cylinder with large radius \(R\). Write your answer in terms of the charge per unit length of the cylinder, which is \(\lambda=\rho \pi a^{2}\). (See Exercise \(1.83\) for a different method of solving this problem.)

Field from a semicircle * A thin plastic rod bent into a semicircle of radius \(R\) has a charge \(Q\) distributed uniformly over its length. Find the electric field at the center of the semicircle.

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