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Chapter 22: Problem 52

An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

Short Answer

Expert verified
(a) The speed of the electron is 53,170 m/s; (b) it travels approximately 66 nm.

Step by step solution

01

Identify Known Variables

We are given that the initial velocity \( v_i = 40 \text{ km/s} = 40,000 \text{ m/s} \), the electric field \( E = 50 \text{ N/C} \), and the time \( t = 1.5 \text{ ns} = 1.5 \times 10^{-9} \text{ s} \). The charge of an electron \( e = 1.6 \times 10^{-19} \text{ C} \), and the mass of an electron \( m = 9.11 \times 10^{-31} \text{ kg} \).
02

Calculate the Force on the Electron

The force on the electron is calculated using the equation \( F = eE \). Substituting in the values, we get: \( F = (1.6 \times 10^{-19} \text{ C})(50 \text{ N/C}) = 8 \times 10^{-18} \text{ N} \).
03

Determine the Acceleration of the Electron

Using Newton's second law, \( F = ma \), we calculate the acceleration \( a \) by rearrranging the formula: \( a = \frac{F}{m} \). Substituting the known values: \( a = \frac{8 \times 10^{-18} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} = 8.78 \times 10^{12} \text{ m/s}^2 \).
04

Calculate the Final Velocity of the Electron

Using the equation for final velocity, \( v_f = v_i + at \), we substitute the known values: \( v_f = 40,000 \text{ m/s} + (8.78 \times 10^{12} \text{ m/s}^2)(1.5 \times 10^{-9} \text{ s}) \). This gives us \( v_f = 53,170 \text{ m/s} \).
05

Calculate the Displacement of the Electron

To find the displacement, use the equation \( d = v_i t + \frac{1}{2} a t^2 \). Substituting the known values: \( d = (40,000 \text{ m/s})(1.5 \times 10^{-9} \text{ s}) + \frac{1}{2} (8.78 \times 10^{12} \text{ m/s}^2)(1.5 \times 10^{-9} \text{ s})^2 \). Solving this gives \( d \approx 66 \text{ nm} \).

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Key Concepts

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

Electric Fields
An electric field is a region around a charged particle where a force is exerted on other charged particles. It is similar to the gravitational field that acts on masses. However, while the gravitational field only attracts, electric fields can either attract or repel, depending on the charges.

In our scenario, the electron enters a uniform electric field, and its motion is influenced by the field鈥檚 direction and magnitude. The electric field strength, given as 50 N/C, tells us the force experienced by a 1-coulomb charge in the field.
  • Force Calculation: The force on a charge is determined by the formula: \( F = eE \), where
    • The electron's charge \( e = 1.6 \times 10^{-19} \) C, and
    • Electric field \( E = 50 \) N/C.
    This yields a force on the electron of \( 8 \times 10^{-18} \) N. This force causes the electron to accelerate or change its velocity, as explored in later sections.
Kinematics
Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause the motion. It mainly deals with parameters like velocity, acceleration, and displacement.

In this problem, kinematics helps us determine the electron's final velocity and the distance traveled. We use initial velocity, acceleration, and time to find these quantities.
  • Equations of Motion:
    • The final velocity equation: \( v_f = v_i + at \).
      • Here,
      • \( v_i = 40,000 \text{ m/s} \) is the initial velocity,
      • \( a = 8.78 \times 10^{12} \text{ m/s}^2 \) is the acceleration,
      • \( t = 1.5 \times 10^{-9} \text{ s} \) is the time.
      This gives \( v_f = 53,170 \text{ m/s} \) as the final velocity.

    • For displacement, the formula: \( d = v_i t + \frac{1}{2} a t^2 \) lets us find the distance traveled by the electron in the given time, which calculates to approximately 66 nm.
    The equations elegantly connect various parameters, providing insights into the motion of the electron in the defined electric field region.
Electron Dynamics
Electron dynamics involves understanding how electrons, being subatomic particles, behave under various conditions, such as the influence of electric and magnetic fields. Given their small mass and charge, electrons can reach high velocities quite rapidly under the influence of such fields.

For this scenario:
  • High Acceleration Response:
    • The acceleration \( a = \frac{8 \times 10^{-18} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} \) leads to extremely high acceleration for an electron.
    This tremendous acceleration allows the electron to greatly increase its speed over a very short time span.
  • Challenges in Measurement: Given the speed and tiny mass involved, precisely measuring or observing these electron dynamics requires delicate instrumentation and consideration of quantum effects.
Since electrons move so quickly and change velocity so rapidly in response to electric fields, their dynamics are quite distinct compared to larger, more massive objects.
Newton's Laws
Newton's Laws fundamentally describe the relationship between a body and the forces acting upon it, providing a framework for understanding motion. In our electron motion problem, these laws, particularly the second law, are crucial.

**Application of Newton's Second Law:**
  • Newton's Second Law states that \( F = ma \), meaning the force applied to an object is equal to the mass of that object multiplied by its acceleration.
    • Here, the force from the electric field and the known mass of the electron allows us to calculate its acceleration.
    This is crucial for predicting how quickly the electron's velocity changes over time.
Using this law, we find that even a tiny force can result in large accelerations for particles like electrons because of their minute mass. This law provides the foundation for analyzing how electrostatic forces influence object motion, seamlessly integrating with other concepts like energy changes and dynamics.

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Most popular questions from this chapter

An electron on the axis of an electric dipole is \(25 \mathrm{~nm}\) from the center of the dipole. What is the magnitude of the electrostatic force on the electron if the dipole moment is \(3.6 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} ?\) Assume that \(25 \mathrm{~nm}\) is much larger than the separation of the charged particles that form the dipole.

Two particles, each of positive charge \(q\), are fixed in place on a \(y\) axis, one at \(y=d\) and the other at \(y=-d\). (a) Write an expression that gives the magnitude \(E\) of the net electric field at points on the \(x\) axis given by \(x=\alpha d\). (b) Graph \(E\) versus \(\alpha\) for the range \(0<\) \(\alpha<4\). From the graph, determine the values of \(\alpha\) that give (c) the maximum value of \(E\) and \((\mathrm{d})\) half the maximum value of \(E .\)

A proton and an electron form two corners of an equilateral triangle of side length \(2.0 \times 10^{-6} \mathrm{~m}\). What is the magnitude of the net electric field these two particles produce at the third corner?

A charge of \(20 \mathrm{nC}\) is uniformly distributed along a straight rod of length \(4.0 \mathrm{~m}\) that is bent into a circular arc with a radius of \(2.0 \mathrm{~m}\). What is the magnitude of the electric field at the center of curvature of the arc?

Two particles, each with a charge of magnitude \(12 \mathrm{nC}\), are at two of the vertices of an equilateral triangle with edge length \(2.0 \mathrm{~m}\). What is the magnitude of the electric field at the third vertex if (a) both charges are positive and (b) one charge is positive and the other is negative?
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