/*! 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 3 In the laboratory frame, a proto... [FREE SOLUTION] | 91Ó°ÊÓ

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In the laboratory frame, a proton is at rest at the origin at \(t=0 .\) At that instant a negative pion (charge \(-e\) ) that has been traveling in along the \(x\) axis at a speed of \(0.6 c\) reaches the point \(x=0.01 \mathrm{~cm}\). There are no other charges around. What is the magnitude of the force on the pion? What is the magnitude of the force on the proton? What about Newton's third law? (We're getting a little ahead of ourselves with this last question, but see if you can answer it anyway.)

Short Answer

Expert verified
The magnitude of the force exerted on both the pion and the proton is 0.2304 N. However, the direction of these forces is opposite for each particle as per Newton's Third Law. The force on the pion is directed towards the proton, while the force on the proton is directed towards the pion.

Step by step solution

01

Understand the task

The exercise involves calculating the force acting on both a proton and a pion moving in a laboratory frame. Since the proton is at rest and the pion is moving, it's crucial to consider the charges of the two particles and their relative positions.
02

Calculate the magnitude of the force on the pion

Consider that the force on the pion at the moment it crosses the proton is purely electric since the proton is initially at rest. The electric force can be given by Coulomb's Law: \(F_e = \frac{k|q1*q2|}{r^2}\), where q1 and q2 are the charges of the pion and proton respectively, r is the distance between them, and k is Coulomb's Constant. Substituting values into the equation gives \(F_e = \frac{(9*10^9 N m^2/C^2)*(-1.6*10^{-19} C)*(1.6*10^{-19} C)}{(0.01*10^{-2} m)^2} = -0.2304 N\). The negative sign indicates the force on the pion is directed towards the proton due to the attractive nature of oppositely charged particles.
03

Calculate the magnitude of the force on the proton

Using Newton's third law, the force on the proton (F_p) should be equal and opposite to the force on the pion: \( F_p = -F_e = 0.2304 N \). This means the proton experiences a force of equal magnitude but in the opposite direction relative to pion.
04

Discussing Newton's Third Law

Newton's Third Law states that every action has an equal and opposite reaction. In this scenario, the electric force exerted on the pion due to the presence of the proton is the action. Consequently, the reaction is the equal and opposite electric force exerted back on the proton due to the pion.

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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 electromagnetism that describes the force between two charged particles. It tells us that the electric force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it's given by the formula:
  • \( F_e = \frac{k |q_1 q_2|}{r^2} \)
Here, \( F_e \) represents the electric force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the separation distance, and \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
This law helps us understand how charged particles like protons and pions interact with each other. In our exercise, this interaction is the reason behind the forces each particle experiences as they come close to each other.
Electric Force
Electric force is the interaction experienced by charged objects. It is a vector quantity, which means it has both magnitude and direction. This force can either attract or repel the involved charges depending on their nature. Opposite charges (like a proton and a negative pion) attract, while like charges repel.
When discussing electric forces in our exercise, the proton and pion experience such forces due to Coulomb's Law. This force is what brings them together, as shown by the formula:
  • \( F_e = \frac{k |q_1 q_2|}{r^2} \)
The calculated force was shown to have a magnitude of about \(0.2304 \text{ N}\). The negative sign indicated an attractive force acting towards the other charge due to their differing signs.
Newton's Third Law
Newton's Third Law is a cornerstone of classical mechanics. It states that for every action, there is an equal and opposite reaction. This means if a body A applies a force on body B, body B will apply an equal and opposite force on body A.
In the case of the proton and pion, when the pion feels a force due to the proton, the proton feels an equal force due to the pion, just in the opposite direction. This is why the magnitude of the electric force on the proton is equal to that on the pion, albeit in the opposite direction, adhering to Newton's Third Law principle.
Proton
A proton is a subatomic particle found in the nucleus of an atom. It carries a positive electric charge, denoted as \(+e\) (\(1.6 \times 10^{-19} \text{ C}\)). It plays a vital role in electromagnetism by interacting with other charged particles, such as electrons and pions.
In our current scenario, the proton remains stationary. However, it interacts with the moving pion by exerting and feeling electric force due to their charge and proximity. This interaction is central to understanding the dynamics of charged particles in physics.
Pion
A pion, or pi meson, is a type of subatomic particle with a negative charge (-e). Unlike stable particles such as protons and electrons, pions are unstable and typically found in high-energy processes. In this exercise, a pion comes into proximity with a proton.
The negative charge of the pion provides the basis for the electric force it experiences when near the proton. As the pion travels near the stationary proton, the resultant electric force pulls it towards the proton, embodying the foundational electromagnetic interactions explored in this exercise. This definite interaction helps students grasp the dynamics involved in electric forces according to Coulomb's Law.

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Most popular questions from this chapter

Transatlantic telegraphic cable *? The first telegraphic messages crossed the Atlantic in 1858 , by a cable \(3000 \mathrm{~km}\) long laid between Newfoundland and Ireland. The conductor in this cable consisted of seven copper wires, each of diameter \(0.73 \mathrm{~mm}\), bundled together and surrounded by an insulating sheath. (a) Calculate the resistance of the conductor. Use \(3.10^{-8}\) ohmmeter for the resistivity of the copper, which was of somewhat dubious purity. (b) A return path for the current was provided by the ocean itself. Given that the resistivity of seawater is about \(0.25\) ohm-meter, see if you can show that the resistance of the ocean return would have been much smaller than that of the cable. (Assume that the electrodes immersed in the water were spheres with radius, say, \(10 \mathrm{~cm}\).)

A capacitor consists of two coaxial cylinders of length \(L\), with outer and inner radii \(a\) and \(b .\) Assume \(L \gg a-b\), so that end corrections may be neglected. Show that the capacitance is \(C=\) \(2 \pi \epsilon_{0} L / \ln (a / b)\). Verify that if the gap between the cylinders, \(a-b\), is very small compared with the radius, this result reduces to onethat could have been obtained by using the formula for the parallelplate capacitor.

A wire above the earth By solving the problem of the point charge and the plane conductor, we have, in effect, solved every problem that can be constructed from it by superposition. For instance, suppose we have a straight wire 200 meters long, uniformly charged with \(10^{-5} \mathrm{C}\) per meter of length, running parallel to the earth at a height of 5 meters. What is the field strength at the surface of the earth, immediately below the wire? (For steady fields the earth behaves like a good conductor.) You may work in the approximation where the length of the wire is much greater than its height. What is the electrical force acting on the wire?

Image charges for two planes : A point charge \(q\) is located between two parallel infinite conducting planes, a distance \(d\) from one and \(\ell-d\) from the other. Where should image charges be located so that the electric field is everywhere perpendicular to the planes?

Two capacitors with the same capacitance \(C\) and charge \(Q\) are placed next to each other, as shown in Fig. 3.39. The two positive plates are then connected by a wire. Will charge flow in the wire? Consider two possible reasonings: (A) Before the plates are connected, the potential differences of the two capacitors are the same (because \(Q\) and \(C\) are the same). So the potentials of the two positive plates are equal. Therefore, no charge will flow in the wire when the plates are connected. (B) Number the plates 1 through 4 , from left to right. Before the plates are connected, there is zero electric field in the region between the capacitors, so plate 3 must be at the same potential as plate 2 . But plate 2 is at a lower potential than plate \(1 .\) Therefore, plate 3 is at a lower potential than plate 1 , so charge will flow in the wire when the plates are connected. Which reasoning is correct, and what is wrong with the wrong reasoning?

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