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Oscillating on a line ** Two positive point charges \(Q\) are located at points \((\pm \ell, 0) .\) A particle with positive charge \(q\) and mass \(m\) is initially located midway between them and is then given a tiny kick. If it is constrained to move along the line joining the two charges \(Q\), show that it undergoes simple harmonic motion (for small oscillations), and find the frequency.

Short Answer

Expert verified
The charge q undergoes simple harmonic motion with a frequency of \(\frac{1}{2\pi} \sqrt{\frac{4K\frac{Qq}{l^3}}{m}}\).

Step by step solution

01

Identify the Forces

First thing is to identify the forces acting on the charge. The charge q feels equal and opposite forces from the two charges Q which are located symmetrically placed at a distance 2l from each other. As per Coulomb's law, the electric force exerted by a charge Q on our charge q can be given by \(F = K \frac{{Qq}}{{r^2}}\) .
02

Apply Newton's Second Law

Since the forces due to both charges Q are equal and opposite, they reinforce in the center and oppose at the ends. This generates a restoring force whenever the test charge q deviates from the center. Now, we can apply the Newton second law which states that force equals mass times acceleration. The force exerted on q is the derivative of the potential energy U with respect to position: \(F = - \frac{dU}{dx}\) . From this, we can derive the expression for F: \( F = -4K \frac{ Qq} {l^3}x\).
03

Relate to Simple Harmonic Motion

The motion is one dimensional and the force is proportional to displacement but with a negative sign indicating it's a restoring force. This is the same form as Hooke's law for springs, \(F=-kx\), where \(k\) is the spring constant. This implies that the motion is an harmonic motion. Thus: \(4K\frac{Qq}{l^3} = k\).
04

Find the Frequency

For simple harmonic motion, the frequency \(\nu\) of the oscillation is given by \(\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\). Substitute k determined from previous step into this formula, we get: \(\nu =\frac{1}{2\pi} \sqrt{\frac{4K\frac{Qq}{l^3}}{m}}\).

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Key Concepts

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

Coulomb's Law
To grasp the concept of simple harmonic motion in charged particles, we first need to understand Coulomb's Law. This fundamental principle in electrostatics describes how the force between two stationary, electrically charged particles relates to the distance between them. Mathematically, it's expressed as
\( F = K \frac{{Qq}}{{r^2}} \),
where \(F\) is the force between the charges, \(K\) is Coulomb's constant, \(Q\) and \(q\) are the magnitudes of the charges, and \(r\) is the distance separating them. In this scenario, since the charged particle is affected by the forces from two fixed charges at equal distances, Coulomb's law helps us determine the initial forces acting on the particle before it starts oscillating.
Newton's Second Law
Understanding the role of Newton's Second Law is crucial when exploring the motion of particles. According to this law, the force acting on an object is equal to the mass of the object times its acceleration, formalized as
\( F = m \times a \).
In our exercise, we applied this law to determine the acceleration of the charged particle. The fact that the forces from both fixed charges reinforce at the center and oppose at the ends results in a restoring force that nudges the particle back toward equilibrium. By setting the electrostatic force equal to mass times acceleration, we're setting up the stage for the particle's oscillatory motion.
Hooke's Law
Hooke's Law provides a link between the forces we've been discussing and simple harmonic motion. This law is commonly associated with springs and states that the force needed to extend or compress a spring by some distance is proportional to that distance, that is
\( F=-kx \),
where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement from the spring's equilibrium position. In the context of our charged particle, the restoring force follows a similar pattern, indicating that our particle indeed undergoes simple harmonic motion. Identifying the electric force as analogous to a spring's restoring force allows us to treat the particle's movement like that of a mass on a spring.
Oscillation Frequency
The oscillation frequency of an object in simple harmonic motion is a pivotal variable that tells us how many oscillations occur per unit of time. For a mass-spring system, or in our case, a charge moving under conditions analogous to Hooke's Law, the frequency is determined by
\( u = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \),
where \(u\) is the frequency, \(k\) is the effective force constant (analogous to the spring constant in Hooke's Law), and \(m\) is the mass of the oscillating object. By calculating the effective force constant from Coulomb's law and incorporating the mass of our charged particle, we deduced the oscillation frequency for the particle constrained to one-dimensional motion between the two fixed charges.

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Most popular questions from this chapter

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.

Zero field \(?\) Four charges, \(q,-q, q\), and \(-q\), are located at equally spaced intervals on the \(x\) axis. Their \(x\) values are \(-3 a,-a, a\), and \(3 a\), respectively. Does there exist a point on the \(y\) axis for which the electric field is zero? If so, find the \(y\) value.

Stable equilibrium in electron jelly ** The task of Exercise \(1.77\) is to find the equilibrium positions of two protons located inside a sphere of electron jelly with total charge \(-2 e\). Show that the equilibria are stable. That is, show that a displacement in any direction will result in a force directed back toward the equilibrium position. (There is no need to know the exact locations of the equilibria, so you can solve this problem without solving Exercise \(1.77\) first.)

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.

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?

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