/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 CP Two positive point charges \(... [FREE SOLUTION] | 91Ó°ÊÓ

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CP 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 \(|x| << a\) . The charge \(q,\) which is free to move along the \(x\) -axis, is then released. (a) Find the frequency of oscillation of the charge \(q .\) (Hint: 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 \(|y| < < a\) , and then released. If this charge is free to move anywhere in the \(x y\) -plane, what will happen to it? Explain your answer.

Short Answer

Expert verified
(a) The frequency of oscillation is \(f = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}}\). (b) The charge will oscillate vertically along the y-axis due to the net restoring force.

Step by step solution

01

Understand the Problem Setup

We have two fixed charges, each with charge \(Q\), located on the \(x\)-axis at positions \(x = a\) and \(x = -a\). Another charge \(q\) (mass \(m\)) is placed at a position \(x\) on the \(x\)-axis and has small oscillations. We will first solve for the frequency of the oscillation when \(|x| << a\).
02

Write the Expression for Electric Force

The electric force between point charges is given by Coulomb's law: \(F = k_e \frac{q_1 q_2}{r^2}\), where \(k_e\) is Coulomb's constant.For charge \(q\) at position \(x\), the forces due to charges at \(a\) and \(-a\) are: \[F_{1} = k_e \frac{Qq}{(a-x)^2}\] and \[F_{2} = k_e \frac{Qq}{(a+x)^2}\].
03

Approximate Using Binomial Expansion

Since \(|x| << a\), we can use the binomial expansion:\[(1 \pm z)^{-2} \approx 1 \mp 2z \]Apply this to the denominators:\[(a-x)^{-2} \approx a^{-2}(1 + 2\frac{x}{a})\]\[(a+x)^{-2} \approx a^{-2}(1 - 2\frac{x}{a})\]
04

Compute Net Force and Simplify

The net force on \(q\) is:\[F_{net} = F_1 - F_2 = k_e Qa^{-2}[1 + 2\frac{x}{a}]q - k_e Qa^{-2}[1 - 2\frac{x}{a}]q\]Simplify to get: \[F_{net} = 4k_e \frac{Qq}{a^3}x\].This is of the form \(F = -kx\), which indicates simple harmonic motion (SHM).
05

Find the Frequency from SHM Characteristics

For SHM, the restoring force is \(F = -kx\) with \(k = 4k_e \frac{Qq}{a^3}\). The angular frequency \(\omega\) is given by:\[\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4k_e Qq}{a^3 m}}\].Thus, the frequency \(f\) is:\[f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}}\].
06

Address the Condition when Charge is on y-axis

When the charge \(q\) is placed on the \(y\)-axis at \(y\), its distances from the charges at \(a\) and \(-a\) are equal. The horizontal components of force due to each fixed charge are equal and opposite, so they cancel out leaving only vertical force components. These lead to net vertical forces, which when combined, result in restoring motion towards \(x=0\), hence leading to oscillations along the \(y\)-axis.

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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 that describes the force between two electric charges. It states that the electric force (F) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The mathematical form of Coulomb's law is given as: - \( F = k_e \frac{q_1 q_2}{r^2} \) Here, \( k_e \) represents Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. Coulomb's law is critical for understanding interactions between charged particles. It's similar to Newton's law of gravitation but applies to electrical forces.
Electric force equations
To analyze how electric charges interact, we use electric force equations based on Coulomb's law. For a scenario with charges fixed on the x-axis and a third charge free to oscillate, the initial forces can be determined using: - \( F_1 = k_e \frac{Qq}{(a-x)^2} \) - \( F_2 = k_e \frac{Qq}{(a+x)^2} \) These equations calculate the electric force exerted by charges positioned at \( x=a \) and \( x=-a \) on a charge \( q \) at position \( x \). The formula involves substituting the respective distances into Coulomb's law to determine the individual forces.
Binomial expansion
The binomial expansion is a mathematical tool that simplifies expressions involving powers. It is especially useful when the expressions have small values in the denominators, allowing easier calculations. In this scenario, since we have \(|x| << a\), small terms in expressions can be expanded using: - \((1 \, \pm \, z)^{-2} \approx 1 \, \mp \, 2z\) For the electric forces given earlier: - \((a-x)^{-2} \approx a^{-2}(1 \, + \, 2\frac{x}{a})\) - \((a+x)^{-2} \approx a^{-2}(1 \, - \, 2\frac{x}{a})\) The binomial expansion simplifies the complexity of expressions, making it easier to derive equations necessary for calculating oscillations.
Oscillation frequency
The oscillation frequency of an object describes how fast it moves back and forth in its repetitive movement, a concept tied to simple harmonic motion (SHM). Here, the charge \( q \) oscillates due to the electric forces between it and the fixed charges. We derive the oscillation frequency from the force equation: - The net force simplifies to: \( F_{net} = 4k_e \frac{Qq}{a^3}x \) - For SHM, the force being proportional to displacement \((-kx)\) indicates that the motion follows a periodic path. The effective spring constant \( k \) here involves electrical terms. the angular frequency \( \omega \) of the charge is: - \( \omega = \sqrt{\frac{4k_e Qq}{a^3 m}} \) The actual frequency \( f \) is given by dividing this by \( 2\pi \): - \( f = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}} \) This describes how fast \( q \) oscillates about its equilibrium position.

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Most popular questions from this chapter

CP Two identical spheres are each attached to silk threads of length \(L=0.500 \mathrm{m}\) and hung from a common point (Fig. P21.68). Each sphere has mass \(m=8.00 \mathrm{g} .\) The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1},\) and the other a different positive charge \(q_{2} ;\) this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\) Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{Cis~placed}\) at the point \(x=+0.150 \mathrm{m}, y=0\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m}\)

A ring-shaped conductor with radius \(a=2.50 \mathrm{cm}\) has a total positive charge \(Q=+0.125 \mathrm{nC}\) uniformly distributed around it, as shown in Fig. \(21.23 .\) The center of the ring is at the origin of coordinates \(O .\) (a) What is the electric field (magnitude and direction) at point \(P,\) which is on the \(x\) -axis at \(x=40.0 \mathrm{cm}\) ? (b) A point charge \(q=-2.50 \mu C\) is placed at the point \(P\) described in part (a). What are the magnitude and direction of the force exerted by the charge \(q\) on the ring?

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

Two particles having charges \(q_{1}=0.500 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC}\) are separated by a distance of 1.20 \(\mathrm{m} .\) At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

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