### Problem StatementA proton enters a region of uniform electric field of magnitude 79.9 N/C with an initial velocity of 20.9 km/s directed perpendicularly to the electric field. What is the speed of the proton 1.93 μs after entering this region?### Options- **20.9 km/s**- **33.3 km/s**- **28.2 km/s**- **30.7 km/s**- **25.6 km/s**### AnalysisThis question involves calculating the speed of a proton after it has interacted with a uniform electric field for a given amount of time. This is a typical problem in electromagnetism where the motion of a charged particle in an electric field is analyzed.### Solution Strategy1. **Initial Conditions:** - Initial velocity \( v_0 = 20.9 \, \text{km/s} \) - Electric field \( E = 79.9 \, \text{N/C} \) - Time \( t = 1.93 \, \mu \text{s} = 1.93 \times 10^{-6} \, \text{s} \)2. **Electric Force on Proton:** \[ F = qE \] Here, \( q \) is the charge of the proton (\( q \approx 1.6 \times 10^{-19} \, \text{C} \)).3. **Acceleration of Proton:** Using Newton's second law, \( F = ma \): \[ a = \frac{F}{m} = \frac{qE}{m} \] Here, \( m \) is the mass of the proton (\( m \approx 1.67 \times 10^{-27} \, \text{kg} \)).4. **Velocity Component Along Electric Field:** The change in velocity due to the electric field over the time interval can be calculated as: \[ \Delta v = at = \frac{qE}{m} \cdot t \]5. **Total Speed of the Proton:** The proton's velocity components are perpendicular to each other, so use Pythagoras’ theorem to find the resultant speed.

Answered: Image /qna-images/answer/09505797-ff99-4309-a7bf-570052a93dfa.jpg

A proton enters a region of uniform electric field of magnitude 79.9 N/C with an initial velocity of 20.9 km/s directed perpendicularly to the electric field. What is the speed of the proton 1.93 μs after entering this region?

A proton enters a region of uniform electric field of magnitude 79.9 N/C with an initial velocity of 20.9 km/s directed perpendicularly to the electric field. What is the speed of the proton 1.93 μs after entering this region?

College Physics

11th Edition

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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One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

### Problem Statement

A proton enters a region of uniform electric field of magnitude 79.9 N/C with an initial velocity of 20.9 km/s directed perpendicularly to the electric field. What is the speed of the proton 1.93 μs after entering this region?

### Options

- **20.9 km/s**
- **33.3 km/s**
- **28.2 km/s**
- **30.7 km/s**
- **25.6 km/s**

### Analysis

This question involves calculating the speed of a proton after it has interacted with a uniform electric field for a given amount of time. This is a typical problem in electromagnetism where the motion of a charged particle in an electric field is analyzed.

### Solution Strategy

1. **Initial Conditions:**
- Initial velocity \( v_0 = 20.9 \, \text{km/s} \)
- Electric field \( E = 79.9 \, \text{N/C} \)
- Time \( t = 1.93 \, \mu \text{s} = 1.93 \times 10^{-6} \, \text{s} \)

2. **Electric Force on Proton:**
\[ F = qE \]
Here, \( q \) is the charge of the proton (\( q \approx 1.6 \times 10^{-19} \, \text{C} \)).

3. **Acceleration of Proton:**
Using Newton's second law, \( F = ma \):
\[ a = \frac{F}{m} = \frac{qE}{m} \]
Here, \( m \) is the mass of the proton (\( m \approx 1.67 \times 10^{-27} \, \text{kg} \)).

4. **Velocity Component Along Electric Field:**
The change in velocity due to the electric field over the time interval can be calculated as:
\[ \Delta v = at = \frac{qE}{m} \cdot t \]

5. **Total Speed of the Proton:**
The proton's velocity components are perpendicular to each other, so use Pythagoras’ theorem to find the resultant speed.

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