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Chapter 24: Problem 47

An electron \(\left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)\) is projected out along the \(+x\) -axis in vacuum with an initial speed of \(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). It goes \(45 \mathrm{~cm}\) and stops due to a uniform electric field in the region. Find the magnitude and direction of the field.

Short Answer

Expert verified
The electric field magnitude is approximately 5.6875 N/C in the -x direction.

Step by step solution

01

Understanding the Problem

We are given an electron moving along the +x-axis that comes to a stop due to a uniform electric field. We need to find the magnitude and direction of this field.
02

Setup the Known Values

We have the initial speed of the electron, \(v_0 = 3.0 \times 10^6 \text{ m/s}\), the final speed \(v = 0\), the distance \(d = 0.45 \text{ m}\), and the mass of the electron \(m_e = 9.1 \times 10^{-31} \text{ kg}\).
03

Using Kinematic Equation to Find Acceleration

We can use the kinematic equation \(v^2 = v_0^2 + 2ad\), where \(v = 0\). Solving for \(a\), we get \(a = -\frac{v_0^2}{2d}\).
04

Calculate the Acceleration

Substitute the known values: \(a = -\frac{(3.0 \times 10^6)^2}{2 \times 0.45}\). Evaluating this, we find \(a = -10^{13} \text{ m/s}^2\).
05

Relate Acceleration to Electric Field

The force acting on the electron is given by \(F = ma\), and the electric force is \(F = qE\), where \(q = -e\). Thus, \(-eE = ma\), leading to \(E = -\frac{ma}{e}\).
06

Calculate the Electric Field

Substitute the known values: \[ E = -\frac{(9.1 \times 10^{-31}) (-10^{13})}{1.6 \times 10^{-19}} \]. Solve to get \(E = 5.6875 \times 10^{-3} \text{ N/C}\).
07

Determine the Direction of the Field

Since the electron is negatively charged and the field stops it, the field is in the opposite direction of the initial velocity, therefore along the -x axis.

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Key Concepts

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

Kinematic Equations
Kinematic equations help us understand motion by relating various physical quantities, such as initial velocity, final velocity, distance, acceleration, and time. In physics, these equations are especially useful when dealing with constant acceleration scenarios. When an electron is subjected to a uniform electric field, its motion is commonly analyzed using these kinematic equations.
One of the essential kinematic equations used in this problem is \(v^2 = v_0^2 + 2ad\), where:
  • \(v\) is the final velocity.
  • \(v_0\) is the initial velocity.
  • \(a\) is the acceleration.
  • \(d\) is the displacement.
In this situation, the electron comes to a stop, which provides us with a final velocity \(v = 0\). By substituting the known initial conditions, the equation becomes a powerful tool to find the missing acceleration value. Understanding how these variables interact is crucial for determining how an electron behaves when subjected to external forces.
Electron Motion
The motion of an electron becomes an intriguing topic when external forces, like an electric field, come into play. Electrons, as negatively charged particles, tend to move toward positive charges within an electric field.
In the given problem, the electron begins moving with a high speed in a specific direction鈥攁long the +x-axis鈥攚ith its motion affected by the electric field. Here are some key points regarding electron motion in this context:
  • The initial velocity direction is due to its initial kinetic energy imparted from an external source.
  • An electron facing opposition by an electric field will experience deceleration.
  • Electric fields influence the electron by applying force according to the charge's polarity; opposites attract, and likes repel.
The field acts on the electron by decelerating it until it stops, which is where we apply kinematic principles to find critical aspects like acceleration and field strength.
Acceleration Calculation
Acceleration is a fundamental concept when analyzing the electron's motion, as it tells us how the electron's velocity changes over time because of the electric field. Here, the acceleration was needed to find the electric field's magnitude.
The calculation involves rearranging the kinematic equation to solve for acceleration \(a\). We know:
  • Initial velocity \(v_0 = 3.0 \times 10^6\) m/s
  • Final velocity \(v = 0\)
  • Distance \(d = 0.45\) m
The acceleration is then calculated as:\[a = -\frac{v_0^2}{2d}\]Substituting the values gives an acceleration of approximately \(-10^{13}\) m/s虏.
This large negative value signifies that the electron is heavily decelerated by the opposing electric field. Understanding how acceleration is calculated helps in determining the electric field's influence, as the relationship between force, charge, and electric field (\(F = ma\), and \(F = qE\)) is key to solving the problem.

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Most popular questions from this chapter

Two small spheres in vacuum are \(1.5 \mathrm{~m}\) apart center-to-center. They carry identical charges. Approximately how large is the charge on each if each sphere experiences a force of \(2 \mathrm{~N}\) ? The diameters of the spheres are small compared to the \(1.5 \mathrm{~m}\) separation. We may therefore approximate them as point charges. Coulomb's Law, \(F_{E}=k_{0} q_{\cdot 1} q_{\cdot 2} / r^{2}\), leads to $$ q_{.1} q_{.2}=q^{2}=\frac{F_{E} r^{2}}{k_{0}}=\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}=5 \times 10^{-10} \mathrm{C}^{2} $$ from which \(q=2 \times 10^{-5} \mathrm{C}\)

Imagine two separated tiny interacting uniformly charged spheres. What happens to the electrostatic force on each of them if the charge on both is doubled and their separation is also doubled?

Three point charges are placed at the following locations on the \(x\) axis: \(+2.0 \mu \mathrm{C}\) at \(x=0,-3.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm},-5.0 \mu \mathrm{C}\) at \(x=120\) \(\mathrm{cm}\). Find the force \((a)\) on the \(-3.0 \mu \mathrm{C}\) charge, \((b)\) on the \(-5.0 \mu \mathrm{C}\) charge.

Find the ratio of the Coulomb electric force \(F_{E}\) to the gravitational force \(F_{G}\) between two electrons in vacuum. From Coulomb's Law and Newton's Law of gravitation, The electric force is much stronger than the gravitational force. $$ \begin{array}{c} F_{E}=k \frac{q^{2}}{r^{2}} \text { and } F_{G}=G \frac{m^{2}}{r^{2}} \\ \text { Therefore, } \quad \begin{aligned} \frac{F_{E}}{F_{G}} &=\frac{k q_{*}^{2} / r^{2}}{G m^{2} / r^{2}}=\frac{k q_{i}^{2}}{G m^{2}} \\ =& \frac{\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(9.1 \times 10^{-31} \mathrm{~kg}\right)^{2}}=4.2 \times 10^{42} \end{aligned} \end{array} $$

Two identical tiny metal balls carry charges of \(+3 \mathrm{nC}\) and \(-12 \mathrm{nC}\). They are \(3 \mathrm{~m}\) apart in vacuum. (a) Compute the force of attraction. (b) The balls are now touched together and then separated to \(3 \mathrm{~cm}\). Describe the forces on them now.
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