Problem 28 - University Physics with Modern Physics

Question

CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively,

Step-by-Step Solution

Answer

(a) The electric field is approximately 5.69 N/C downward. (b) Yes, gravity's effect is negligible due to its minimal force compared to the electric force.

Step 1: Understanding the Problem

An electron is released from rest in a uniform electric field and travels a certain distance upwards within a given time. We are to find the electric field's magnitude and direction based on the electron's motion.

Step 2: Data Identification

The distance traveled by the electron: \[ s = 4.50 \text{ m} \]The time taken: \[ t = 3.00 \times 10^{-6} \text{ s} \]Initial velocity: \[ u = 0 \text{ (since it is released from rest)} \]

Step 3: Using Kinematic Equation

Apply the equation of motion \[ s = ut + \frac{1}{2}at^2 \] to find acceleration \( a \). Since the initial velocity \( u = 0 \):\[ 4.50 = 0 \times t + \frac{1}{2}a(3.00 \times 10^{-6})^2 \]Solving for \( a \), we find:\[ a = \frac{2 \times 4.50}{(3.00 \times 10^{-6})^2} \approx 10^{12} \text{ m/s}^2 \]

Step 4: Determine Electric Field Strength

The force \( F \) on the electron is given by \( F = ma \), where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \text{ kg} \)). Thus,\[ F = (9.11 \times 10^{-31})(10^{12}) = 9.11 \times 10^{-19} \text{ N} \]The electric field \( E \) is found using \( F = eE \), with \( e \) the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)).\[ E = \frac{F}{e} = \frac{9.11 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 5.69 \text{ N/C} \]

Step 5: Determine Direction of Electric Field

Because the electron is negatively charged and moves upward, the electric field must point downward.

Step 6: Evaluating Gravity's Influence

Calculate the gravitational force on the electron: \[ F_g = mg = (9.11 \times 10^{-31})(9.8) = 8.93 \times 10^{-30} \text{ N} \]Compare with the electric force \( F_e = 9.11 \times 10^{-19} \text{ N} \). The electric force is significantly larger, justifying the neglect of gravity.

Key Concepts

Kinematic Equations

Kinematic equations are fundamental tools in physics used to describe the motion of objects. These equations relate the initial velocity, final velocity, acceleration, time, and displacement of an object. In this exercise, we used the kinematic equation:

• \[ s = ut + \frac{1}{2}at^2 \]

Here:

• \( s \) is the displacement, 4.50 meters in this case.
• \( u \) is the initial velocity, which is 0 as the electron is released from rest.
• \( a \) is the acceleration we want to find.
• \( t \) is the time, 3 microseconds or \( 3.00 \times 10^{-6} \) seconds.

By solving for \( a \), we determine the electron's acceleration due to the electric field. These equations allow us to calculate one missing variable if the others are known, proving essential in various physical contexts, such as projectile motion and free-fall calculations.
Understanding kinematic equations helps in predicting the behavior of moving objects under uniform acceleration, which is crucial in analyzing electronic and gravitational systems.

Electron Acceleration

When we're discussing the acceleration of an electron in an electric field, it is critical to understand that different forces are at play. Electrons have a small mass and are subject to electromagnetic forces when in an electric field. The equation used here is:

• \[ F = ma \]

For an electron, \( m \) is its mass, and \( a \) is its acceleration due to the force exerted by the electric field.

• The mass of an electron is \( 9.11 \times 10^{-31} \) kg, and the acceleration due to the electric field was calculated as \( 10^{12} \) m/s².
• Using Newton’s second law, \( F = ma \), we determine that the force on the electron is \( 9.11 \times 10^{-19} \) Newtons.

This demonstrates how effectively electric fields can accelerate charged particles due to their small masses, significant in fields like electronics and particle physics. Electrons can gain high velocities over short distances, explaining their rapid movements in circuits and devices.

Gravity Influence

Gravity is a force that usually acts on objects with mass, pulling them toward the Earth's center. However, in this exercise, gravity's influence is minimal compared to the electric force acting on the electron.

• The gravitational force \( F_g \) acting on the electron is calculated as:
• \[ F_g = mg \]
• Where \( m \) is the electron's mass \( (9.11 \times 10^{-31} \text{ kg}) \) and \( g \) is the acceleration due to gravity \( (9.8 \text{ m/s}^2) \).

This results in a gravitational force of \( 8.93 \times 10^{-30} \text{ N} \), which is significantly smaller than the electric force of \( 9.11 \times 10^{-19} \text{ N} \).
Given the vast difference in magnitudes, it’s clear why the gravitational force is negligible in this context. The dominance of the electric force means gravity's impact on an electron in a strong electric field is minimal, allowing us to focus primarily on electric interactions in many physical applications.

Electric Charge

Electric charge is a fundamental property of matter, existing in two types: positive and negative. Electrons, with a negative charge, are greatly influenced by electric fields. The interaction between charge and electric fields is described by:

• \[ F = eE \]

Here, \( e \) is the electron's charge \( (1.6 \times 10^{-19} \text{ C}) \), and \( E \) is the strength of the electric field. From this exercise:

• Given the electric force \( F \) acting on the electron is \( 9.11 \times 10^{-19} \) N, we calculate:
• \[ E = \frac{F}{e} \approx 5.69 \text{ N/C} \]

This calculation shows the relation between the force experienced by a charged particle and the electric field's intensity. Additionally, because electrons repel each other and are attracted to positive charges, the electron's upward movement in this scenario indicates a downward-pointing electric field.
Understanding electric charge and its interaction with fields is fundamental in electromagnetism, is crucial for designing electric circuits, and plays a vital role in various technologies such as capacitors, transistors, and batteries.