An electron is released from rest in a uniform electric field of magnitude 1.58 x 104 N/C. Calculate the acceleration of the electron. (Ignore gravitation.) Number i Units

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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### Problem Description

An electron is released from rest in a uniform electric field of magnitude \(1.58 \times 10^4 \, \text{N/C}\). Calculate the acceleration of the electron. (Ignore gravitation.)

### Data Entry

- **Number**: _______
- **Units**: _______

### Explanation

In this problem, students are required to calculate the acceleration of an electron when it is released from rest in a uniform electric field. The given electric field magnitude is \(1.58 \times 10^4 \, \text{N/C}\). Gravitational effects are to be ignored for this calculation.

To solve this problem, students need to understand the relationship between the electric force and acceleration, using Newton's Second Law of Motion.

**Hint:**
1. The force on an electron in an electric field is given by \( F = eE \).
   - \( e \) is the elementary charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)).
   - \( E \) is the electric field strength.
2. Use Newton's Second Law \( F = ma \) to solve for acceleration \( a \).
   - \( m \) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)).
   - \( a \) is what we need to find.

### Solution Submission

Students should enter their numerical answer in the "Number" field and select the appropriate units from the drop-down menu in the "Units" field.
Transcribed Image Text:### Problem Description An electron is released from rest in a uniform electric field of magnitude \(1.58 \times 10^4 \, \text{N/C}\). Calculate the acceleration of the electron. (Ignore gravitation.) ### Data Entry - **Number**: _______ - **Units**: _______ ### Explanation In this problem, students are required to calculate the acceleration of an electron when it is released from rest in a uniform electric field. The given electric field magnitude is \(1.58 \times 10^4 \, \text{N/C}\). Gravitational effects are to be ignored for this calculation. To solve this problem, students need to understand the relationship between the electric force and acceleration, using Newton's Second Law of Motion. **Hint:** 1. The force on an electron in an electric field is given by \( F = eE \). - \( e \) is the elementary charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)). - \( E \) is the electric field strength. 2. Use Newton's Second Law \( F = ma \) to solve for acceleration \( a \). - \( m \) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)). - \( a \) is what we need to find. ### Solution Submission Students should enter their numerical answer in the "Number" field and select the appropriate units from the drop-down menu in the "Units" field.
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