/*! 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 52 An electron is accelerated eastw... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron is accelerated eastward at \(3.40 \times 10^{9} \mathrm{~m} / \mathrm{s}^{2}\) by an electric field. Determine the field (a) magnitude and (b) direction.

Short Answer

Expert verified
(a) 1.93 N/C; (b) The electric field is westward.

Step by step solution

01

Understanding the formula

To determine the electric field, we need to use the formula that relates acceleration, charge, and electric field, which is given by Newton's second law: \[ F = ma \] where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In the context of electric fields, the force on a charge \( q \) is also given by \( F = qE \), where \( E \) is the electric field. Therefore, \[ qE = ma \] and we can solve for \( E \) by rearranging to \[ E = \frac{ma}{q} \].
02

Inserting known values

To proceed, we need the mass and charge of an electron. These values are: \( m = 9.11 \times 10^{-31} \) kg (mass of an electron) and \( q = 1.60 \times 10^{-19} \) C (charge of an electron). We are given \( a = 3.40 \times 10^{9} \) m/s extsuperscript{2}. Insert these values into the equation: \[ E = \frac{(9.11 \times 10^{-31} \, \text{kg})(3.40 \times 10^{9} \, \text{m/s}^2)}{1.60 \times 10^{-19} \, \text{C}} \].
03

Calculating the magnitude of the electric field

Now, calculate the magnitude of the electric field by performing the multiplication and division: \[ E = \frac{3.40 \times 9.11 \times 10^{-22}}{1.60 \times 10^{-19}} \] This simplifies to \[ E \approx \frac{30.974 \times 10^{-22}}{1.60 \times 10^{-19}} \]. Finally, divide the numbers to obtain \[ E \approx 1.93 \times 10^{-2} \text{ N/C} \].
04

Determining the direction of the electric field

Since the electron is accelerated eastward and electrons have a negative charge, the electric field that causes the acceleration must be directed in the opposite direction of the electron's acceleration to account for its negative charge. Therefore, the electric field direction is westward.

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly the velocity of an object changes with time. It occurs whenever there is a change in speed or direction. In the case of linear motion, acceleration is calculated by dividing the change in velocity by the time over which the change occurs. When you know the acceleration of an object, it can tell you how fast that object will speed up or slow down when acted upon by a force.
For the exercise at hand, the given acceleration of the electron is extremely high, approximately \[ 3.40 \times 10^9 \, \text{m/s}^2 \]. This means that the electron's velocity is increasing incredibly fast in a very short span of time. Understanding this large value is crucial when considering the forces required to produce such acceleration, which ties directly into the concepts of electric fields and forces.
In context, the acceleration direction—as eastward—helps in determining how the electric field interacts with the electron, an aspect crucial for solving the original problem.
Electron
The electron is a subatomic particle with a fundamental negative electric charge of \(- 1.60 \times 10^{-19} \, \text{C} \). Electrons are significantly lighter than protons and neutrons, with a mass of roughly \(9.11 \times 10^{-31} \, \text{kg} \). These tiny particles are crucial in electricity, electronics, and many other physical phenomena.
In electrical contexts, understanding the role of electrons is vital since they carry electric current in most materials. They respond to electric fields, which can cause them to accelerate or change direction, depending on the field's properties.
In the exercise problem, the electron acts as the charged particle affected by the electric field, which provides both a real-world application of the discussed theory and insight into how electric fields can be used to manipulate these particles.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle that connects the net force acting on an object to its mass and the resulting acceleration: \[ F = ma \]. This law forms the foundation for analyzing motion, where:
  • \( F \) is the net force in newtons (N),
  • \( m \) is the mass of the object in kilograms (kg),
  • \( a \) is the acceleration in meters per second squared (m/s²).
In this context, when dealing with charged particles, the force can also be related to the electric field and charge: \[ F = qE \], where:
  • \( q \) is the charge of the particle in coulombs (C),
  • \( E \) is the electric field strength in newtons per coulomb (N/C).
For an electron in an electric field, the Second Law helps to determine the electric field's magnitude necessary to produce the given acceleration. It links the fundamental interactions between forces, motion, and fields.
Electric Charge
Electric charge is a property that causes subatomic particles such as electrons and protons to experience a force when placed in an electric field. Charges can be positive or negative, and like charges repel each other while opposite charges attract.
The SI unit of electric charge is the coulomb (C). An electron has a negative charge of \(- 1.60 \times 10^{-19} \, \text{C} \), which is equal in magnitude but opposite in sign to the charge of a proton.
Understanding electric charge is essential for explaining how electric fields interact with matter. In the original exercise, the negative charge of the electron determines the direction of the force it experiences in an electric field. This influences the direction of acceleration when a charged particle is subjected to a field, thereby affecting the calculations involving electric fields and forces.

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Most popular questions from this chapter

Charge is uniformly distributed around a ring of radius \(R=4.60 \mathrm{~cm}\), and the resulting electric field magnitude \(E\) is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is \(E\) maximum?

Figure 22-36 shows two concentric rings, of radii \(R\) and \(R^{\prime}=4.00 R\), that lie on the same plane. Point \(P\) lies on the central z axis, at distance \(D=2.00 R\) from the center of the rings. The smaller ring has uniformly distributed charge \(+Q .\) In terms of \(Q\), what is the uniformly distributed charge on the larger ring if the net electric field at \(P\) is zero?

A thin nonconducting rod with a uniform distribution of positive charge \(Q\) is bent into a complete circle of radius \(R\) (Fig. 22-22). The central perpendicular axis through the ring is a \(z\) axis, with the origin at the center of the ring. What is the magnitude of the electric field due to the rod at (a) \(z=0\) and (b) \(z=\infty\) ? (c) In terms of \(R\), at what positive value of \(z\) is that magnitude maximum? (d) If \(R=2.00 \mathrm{~cm}\) and \(Q=5.00 \mu \mathrm{C}\), what is the maximum magnitude?

Two charged particles are fixed to an \(x\) axis: Particle 1 of charge \(q_{1}=2.1 \times 10^{-8} \mathrm{C}\) is at position \(x=20 \mathrm{~cm}\) and particle 2 of charge \(q_{2}=-4.00 q_{1}\) is at position \(x=70 \mathrm{~cm}\). (a) At what coordinate on the axis (other than at infinity) is the net electric field produced by the two particles equal to zero? (b) What is the zero-field coordinate if the particles are interchanged?

Humid air breaks down (its molecules become ionized) in an electric field of \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\). In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing, and (c) what is the acceleration magnitude of a free electron?

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