Two uncharged spheres are separated by 2.30 m. If 1.70 X 10^12 electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
Two uncharged spheres are separated by 2.30 m. If 1.70 X 10^12 electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
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
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Two uncharged spheres are separated by 2.30 m. If 1.70 X 10^12 electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
![### Physics Problems from SERCP11 (Educational Resource)
#### Problem 3 [0/6 Points]
**Details** | **Previous Answers** | **SERCP11 15.2.P.007** | **1/5 Submissions Used**
**Problem:**
Two uncharged spheres are separated by 2.30 m. If 1.70 x 10^12 electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
**Hint:**
\( N \)
**Need Help?**
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- [Watch It]
#### Problem 4 [6/6 Points]
**Details** | **Previous Answers** | **SERCP11 15.3.QP.020** | **1/5 Submissions Used**
**Problem:**
Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
- **Hints:**
- The calculations involve finding the net force on the electron using \( F = qE \) and then using Newton's second law \( F = ma \) to find the acceleration.
\[ 5.62 \times 10^{13} \, \text{m/s}^2 \] (checked)
\[ 5.62 \times 10^{13} \, \text{m/s}^2 \]
(b) The electron is initially at rest. What is its speed (in m/s) after 1.05 x 10^-8 s?
- **Hints:**
- Use the equation of motion \( v = u + at \), where \( u \) is initial velocity (0 in this case), \( a \) is acceleration, and \( t \) is time.
\[ 5.91 \times 10^5 \, \text{m/s} \] (checked)
\[ 5.90 \times 10^5 \, \text{m/s} \]
**Solution or Explanation:**
- The detailed solution includes intermediate steps displayed for accurate results, even though the calculations are made using unrounded values.
**Need Help?**
- [Read It]
---
These physics problems are designed to test your understanding of Coulomb’s law](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1e92540-cdcd-4061-bb75-104670a077d4%2F5668d5d5-bcad-46fb-91f6-bfbdf577ccc0%2Faxlthe9.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problems from SERCP11 (Educational Resource)
#### Problem 3 [0/6 Points]
**Details** | **Previous Answers** | **SERCP11 15.2.P.007** | **1/5 Submissions Used**
**Problem:**
Two uncharged spheres are separated by 2.30 m. If 1.70 x 10^12 electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
**Hint:**
\( N \)
**Need Help?**
- [Read It]
- [Watch It]
#### Problem 4 [6/6 Points]
**Details** | **Previous Answers** | **SERCP11 15.3.QP.020** | **1/5 Submissions Used**
**Problem:**
Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
- **Hints:**
- The calculations involve finding the net force on the electron using \( F = qE \) and then using Newton's second law \( F = ma \) to find the acceleration.
\[ 5.62 \times 10^{13} \, \text{m/s}^2 \] (checked)
\[ 5.62 \times 10^{13} \, \text{m/s}^2 \]
(b) The electron is initially at rest. What is its speed (in m/s) after 1.05 x 10^-8 s?
- **Hints:**
- Use the equation of motion \( v = u + at \), where \( u \) is initial velocity (0 in this case), \( a \) is acceleration, and \( t \) is time.
\[ 5.91 \times 10^5 \, \text{m/s} \] (checked)
\[ 5.90 \times 10^5 \, \text{m/s} \]
**Solution or Explanation:**
- The detailed solution includes intermediate steps displayed for accurate results, even though the calculations are made using unrounded values.
**Need Help?**
- [Read It]
---
These physics problems are designed to test your understanding of Coulomb’s law
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