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Two positive point charges \(Q\) are held fixed on the \(x\)-axis at \(x=a\) and \(x=-a\). A third positive point charge \(q\), with mass \(m\), is placed on the \(x\)-axis away from the origin at a coordinate \(x\) such that \(|x| \ll a\). The charge \(q\), which is free to move along the \(x\)-axis, is then released. (a) Find the frequency of oscillation of the charge \(q\). (Himt: Review the definition of simple harmonic motion in Section 14.2. Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots\), valid for the case \(|z|<1 .\) ) (b) Suppose instead that the charge \(q\) were placed on the \(y\)-axis at a coordinate \(y\) such that \(|y|

Short Answer

Expert verified
The oscillation frequency is \( \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{ma^3}} \). For part (b), the charge moves toward the origin on the y-axis.

Step by step solution

01

Analyze the forces on the charge q

Consider the forces acting on the small charge \( q \) placed at a position \( x \) on the x-axis. Each charge \( Q \) exerts a repulsive force on \( q \): the charge at \( x = a \) exerts a force \( F_1 \) and the charge at \( x = -a \) exerts a force \( F_2 \). Since \( |x| \ll a \), use the approximation \( x \approx 0 \) to simplify calculations.
02

Calculate the forces using Coulomb's Law

The force between two charges is given by Coulomb’s law: \( F = \frac{k_e Qq}{r^2} \), where \( r \) is the distance between the charges. Thus, the forces on \( q \) from \( Q \) located at \( x = a \) is \( F_1 = \frac{k_e Q q}{(a-x)^2} \) and from \( Q \) at \( x = -a \) is \( F_2 = \frac{k_e Q q}{(a+x)^2} \).
03

Use binomial expansion for force approximation

Using the binomial expansion \( (1+z)^{-2} \approx 1 - 2z + 3z^2 \) for \( |z| \ll 1 \), express \( \frac{1}{(a \pm x)^2} \approx \frac{1}{a^2}(1 \pm \frac{2x}{a}) \). Substitute these into \( F_1 \) and \( F_2 \).
04

Find net force and show simple harmonic motion

The net force \( F_{net} = F_1 - F_2 = \frac{4k_e Qqx}{a^3} \). This force is proportional to \( -x \), which indicates that the motion is simple harmonic. The equation of motion is \( m \ddot{x} = -\frac{4k_e Qq}{a^3}x \).
05

Determine the frequency of oscillation

For simple harmonic motion, \( m \ddot{x} = -kx \) where \( k = \frac{4k_e Qq}{a^3} \). The angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \), so \( \omega = \sqrt{\frac{4k_e Qq}{ma^3}} \). The frequency of oscillation \( f \) is \( \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{ma^3}} \).
06

Analyze behavior when charge is placed on y-axis

If the charge \( q \) is placed on the y-axis at \( y \), the forces exerted by each \( Q \) would result in a net radial force pulling \( q \) toward the origin, as the components along the y-axis would cancel each other. This results in the charge moving inwards along the y-axis with increasing velocity until it reaches the origin at \( (0,0) \).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics that describes the force between two point charges. It states that the force (F) between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as \( F = \frac{k_e Q_1 Q_2}{r^2} \), where \( k_e \) is Coulomb's constant (approximately \( 8.99 \times 10^9 \text{N m}^2/\text{C}^2 \)), \( Q_1 \) and \( Q_2 \) are the charges, and \( r \) is the distance between the charges.
  • This law explains how charges interact - either repelling or attracting each other.
  • When applying Coulomb's Law, always ensure units are consistent for force, charge, and distance.

In our given problem, each fixed charge \( Q \) at positions \( x = a \) and \( x = -a \) exert a repulsive force on the moving charge \( q \) along the x-axis.
Using symmetry and Coulomb’s Law simplifies calculations, especially when evaluating small displacements along the x-axis.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion where the force acting on an object is proportional to the displacement of the object and is directed towards the object's equilibrium position. This can be simplified into the formula \( F = -kx \) where \( k \) is a constant and \( x \) is the displacement.
  • Important characteristics of SHM include constant amplitude, constant frequency, and sinusoidal wave forms.
  • Analyzing forces in SHM often reveals patterns or symmetries that make complex problems easier to solve.

In our problem, the force calculated between charge \( q \) and the fixed charges \( Q \) is akin to the restoring force in SHM. The net restoring force is proportional to \( -x \), confirming that \( q \) moves in simple harmonic motion along the x-axis.
This harmonic motion makes it possible to predict the charge’s behavior over time.
Binomial Expansion
The Binomial Expansion is an algebraic tool used to simplify calculations involving powers of \((1+z)\), especially when \(|z| \ll 1\). The expansion can be expressed as \((1+z)^n \approx 1 + nz + \frac{n(n-1)}{2}z^2 + \cdots\). The ability to approximate complicated expressions as a series is especially useful in physics.
  • It allows us to make approximations where the value of \(z\) is small, simplifying complex expressions.
  • High-order terms can often be neglected, making computations more tractable.

In the given exercise, using the Binomial Expansion for approximating \(\frac{1}{(a \pm x)^2}\) simplifies the calculation of forces on the charge \( q \).
By expanding to the first order, we can approximate the expressions more easily, which leads us directly to the conclusion about the behavior of the oscillating charge.
Oscillation Frequency
Oscillation Frequency refers to the number of oscillations that happen per unit of time. For an object in simple harmonic motion, frequency is closely related to the systems' physical parameters such as mass and spring constant (or electrostatic forces in this case). The formula used is \( f = \frac{\omega}{2\pi} \), where \( \omega \) is the angular frequency given by \( \sqrt{\frac{k}{m}} \).
  • The angular frequency \( \omega \) itself reflects how quickly an object moves through one complete cycle of motion.
  • Key factors like mass \( m \) and the effective "spring" constant \( k \) (originating from force terms) determine the oscillation frequency.

In this problem, the frequency of charge \( q \)'s oscillation is derived from the equation of motion for SHM (\( m \ddot{x} = -kx \)).
The expression \( \omega = \sqrt{\frac{4k_e Qq}{ma^3}} \) emerges from matching terms, leading directly to the frequency by dividing \( \omega \) by \( 2\pi \).
Examining these derivations clarifies how physical constraints influence the frequency at which the system oscillates.

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Most popular questions from this chapter

Three point charges are placed on the \(y\)-axis: a charge \(q\) at \(y=a\), a charge \(-2 q\) at the origin, and a charge \(q\) at \(y=-a\). Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive \(x\)-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for \(x \gg a\). Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.

Two point charges are located on the \(y\)-axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m}\), and charge \(q_{2}=\) \(+3.20 \mathrm{nC}\) at the origin \((y=0)\). What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu s\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P1.57). 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 \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus in Fig. P1.56. (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.)

A very long, straight wire has charge per unit length \(1.50 \times 10^{-10} \mathrm{C} / \mathrm{m}\). At what distance from the wire is the electricfield magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)

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