Problem 41 - University Physics with Modern Physics

Question

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A ?(\mathrm{b}) \mathrm{A}\) proton is moving in the uniform electric field of part (a). At point \(A .\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s}\) , east. What is the speed of the proton at point \(B ?\)

Step-by-Step Solution

Answer

Electron: \(2.29 \times 10^5\) m/s, Proton: \(2.17 \times 10^4\) m/s.

Step 1: Understanding the Concept

In a uniform electric field, an electron experiences a constant force opposite to the direction of the field. This implies it decelerates when moving in the direction of the electric field (east in this case). The kinetic energy of the electron changes due to the work done by the field.

Step 2: Calculate the Force on the Electron

The force on an electron in an electric field is given by \( F = qE \), where \( q = -1.6 \times 10^{-19} \) C is the charge of an electron and \( E = 1.50 \) N/C is the electric field strength. The force \( F = -1.6 \times 10^{-19} \times 1.50 = -2.4 \times 10^{-19} \) N (opposite to the electric field direction).

Step 3: Determine the Acceleration of the Electron

Using Newton's second law, \( F = ma \), solve for the acceleration \( a = \frac{F}{m} = \frac{-2.4 \times 10^{-19}}{9.11 \times 10^{-31}} \approx -2.63 \times 10^{11} \ \text{m/s}^2 \), where \( m = 9.11 \times 10^{-31} \) kg is the mass of the electron.

Step 4: Use Kinetic Equations to Determine Final Velocity for Electron

Use the equation \( v^2 = u^2 + 2a s \), where \( u = 4.50 \times 10^5 \ \text{m/s} \), \( a = -2.63 \times 10^{11} \ \text{m/s}^2 \), and \( s = 0.375 \ \text{m} \). Solving gives:\[ v^2 = (4.50 \times 10^5)^2 + 2 \times (-2.63 \times 10^{11}) \times 0.375 \]\[ v^2 = 2.025 \times 10^{11} - 1.9725 \times 10^{11} \]\[ v^2 = 0.0525 \times 10^{11} \]\[ v \approx 2.29 \times 10^5 \ \text{m/s} \]

Step 5: Understanding Proton Dynamics

A proton moving in the same electric field will be accelerated in the direction of the field since it has a positive charge. We'll apply similar steps to find its final velocity.

Step 6: Calculate the Force and Acceleration for the Proton

The force on the proton is \( F = qE = 1.6 \times 10^{-19} \times 1.50 = 2.4 \times 10^{-19} \ \text{N} \). The mass of a proton is \( 1.67 \times 10^{-27} \ \text{kg} \). So the acceleration \( a = \frac{2.4 \times 10^{-19}}{1.67 \times 10^{-27}} = 1.44 \times 10^8 \ \text{m/s}^2 \).

Step 7: Determine the Final Velocity for the Proton

Using the equation \( v^2 = u^2 + 2a s \), where \( u = 1.90 \times 10^4 \ \text{m/s} \), \( a = 1.44 \times 10^8 \ \text{m/s}^2 \), and \( s = 0.375 \ \text{m} \). Solve for \( v \):\[ v^2 = (1.90 \times 10^4)^2 + 2 \times 1.44 \times 10^8 \times 0.375 \]\[ v^2 = 3.61 \times 10^8 + 1.08 \times 10^8 \]\[ v^2 = 4.69 \times 10^8 \]\[ v \approx 2.17 \times 10^4 \ \text{m/s} \]

Step 8: Conclusion

The calculations reveal the final speeds of both particles after traveling through the electric field.

Key Concepts

Electron Dynamics

Understanding electron dynamics involves analyzing the behavior of electrons in different environments. In the context of electric fields, electrons, which are negatively charged, will experience forces opposite to the field direction. This is because electric fields exert forces on charges proportional to the field's direction and strength.
For an electron in a uniform electric field like the one described in the problem, this means it decelerates when moving in the direction of the electric field. The force exerted is calculated using the formula:

• The force, \(F\), is given by \(F = qE\), where \(q\) is the electron's charge (\(-1.6 \times 10^{-19}\) C). The result is that the force is opposite the field's direction, resulting in the charge slowing down when moving east in this scenario.

This concept is critical when determining an electron's acceleration, as found via Newton's second law, and influences how its velocity changes as it moves through the field.

Proton Dynamics

Proton dynamics differ from electron dynamics due to the positive charge of protons. In a uniform electric field, protons experience a force in the same direction as the field. This causes them to accelerate when moving along the direction of the field. This scenario leads to an increase in the proton’s speed rather than the deceleration observed in electrons.
Similar to electrons, the force exerted on the proton is calculated using the formula:

• \(F = qE\), with \(q = 1.6 \times 10^{-19}\) C reflecting the positive charge.

Once the force acting on the proton is known, Newton's second law (\(F = ma\)) allows us to determine the proton’s acceleration. The acceleration is then used to compute the increase in velocity as the proton travels through a given distance in the electric field. Understanding proton dynamics is essential for predicting how protons will behave under the influence of various electric fields.

Kinetic Equations

Kinetic equations are invaluable tools when solving problems involving moving charges in electric fields. They help us relate an object’s initial and final velocities, acceleration, and displacement.
In the equations of motion, the form \(v^2 = u^2 + 2as\) is particularly useful as it avoids needing time as a variable. Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement. This equation helps us calculate an electron's or proton’s speed after it has traveled a certain distance under a specific force.

• This approach is used by determining the change in kinetic energy as an object moves through an electric field, accounting for the work done on the charge by the electric field.
• It’s crucial to apply the correct sign for acceleration, reflecting whether the object is accelerating or decelerating, which corresponds to the direction of the force applied.

By mastering these kinetic equations, understanding how charged particles respond to electric fields becomes more intuitive.

Uniform Electric Field

A uniform electric field is one where the electric field strength is constant in magnitude and direction throughout. Imagine a set of evenly spaced parallel lines; the field becomes an environment where charges experience predictable forces.
The simplicity of a uniform electric field allows for straightforward calculations of forces and accelerations applied to any charged particle within the field. The electric field strength, \(E\), is calculated using the ratio of force to charge, \(F = qE\). It remains consistent across any part of the field, simplifying the change in kinetic energy as a charge passes through it.

• For electrons and protons moving in a uniform electric field, like those in the problem, transformations in motion are linear. This means they either accelerate or decelerate steadily.
• This concept is vital for predicting the behavior of charges and conducting effective calculations regarding their dynamics under various electric field conditions.

Grasping the nature of a uniform electric field provides foundational knowledge required to solve complex electric field problems.