How far apart must be a proton from a group of 27 protons (held in place next to one another) if the magnitude of the electrostati force acting on the lone proton due to the group is equal to the magnitude of the gravitational force on the lone proton at Earth's surface? Number i Units
How far apart must be a proton from a group of 27 protons (held in place next to one another) if the magnitude of the electrostati force acting on the lone proton due to the group is equal to the magnitude of the gravitational force on the lone proton at Earth's surface? Number i Units
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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![### Physics Problem: Electrostatic and Gravitational Forces
**Question:**
How far apart must be a proton from a group of 27 protons (held in place next to one another) if the magnitude of the electrostatic force acting on the lone proton due to the group is equal to the magnitude of the gravitational force on the lone proton at Earth's surface?
**Answer:**
- **Number:** [Input box for the answer]
- **Units:** [Dropdown for unit selection]
---
This problem explores the balance between the electrostatic force and gravitational force acting on a proton, providing real-world applications of fundamental physical concepts.
**Details:**
- **Electrostatic Force (Coulomb's Law):** This is the force between charged particles. For protons, this repulsive force can be calculated using Coulomb’s Law:
\[ F_e = k_e \frac{q_1 q_2}{r^2} \]
where \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are charges of the protons, and \( r \) is the distance between the charges.
- **Gravitational Force:** The gravitational force acting on the proton at Earth's surface can be determined by:
\[ F_g = m g \]
where \( m \) is the mass of the proton (\( 1.67 \times 10^{-27} \, \text{kg} \)), and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
- **Solving for Distance:** To find the distance \( r \) such that:
\[ F_e = F_g \]
Rearrange the formulas and solve for \( r \):
\[ k_e \frac{q^2}{r^2} = m g \]
Rearranging for \( r \), we get:
\[ r = \sqrt{\frac{k_e q^2}{m g}} \]
**Assumptions:**
- The group of 27 protons is considered as a point charge situated very closely to one another.
- The lone proton’s interactions are limited to electrostatic](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe462800b-ac16-4590-bafb-82d68c005ec5%2F9d941345-99a4-451b-89a0-74ba8d35d370%2Flzogjon_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Electrostatic and Gravitational Forces
**Question:**
How far apart must be a proton from a group of 27 protons (held in place next to one another) if the magnitude of the electrostatic force acting on the lone proton due to the group is equal to the magnitude of the gravitational force on the lone proton at Earth's surface?
**Answer:**
- **Number:** [Input box for the answer]
- **Units:** [Dropdown for unit selection]
---
This problem explores the balance between the electrostatic force and gravitational force acting on a proton, providing real-world applications of fundamental physical concepts.
**Details:**
- **Electrostatic Force (Coulomb's Law):** This is the force between charged particles. For protons, this repulsive force can be calculated using Coulomb’s Law:
\[ F_e = k_e \frac{q_1 q_2}{r^2} \]
where \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are charges of the protons, and \( r \) is the distance between the charges.
- **Gravitational Force:** The gravitational force acting on the proton at Earth's surface can be determined by:
\[ F_g = m g \]
where \( m \) is the mass of the proton (\( 1.67 \times 10^{-27} \, \text{kg} \)), and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
- **Solving for Distance:** To find the distance \( r \) such that:
\[ F_e = F_g \]
Rearrange the formulas and solve for \( r \):
\[ k_e \frac{q^2}{r^2} = m g \]
Rearranging for \( r \), we get:
\[ r = \sqrt{\frac{k_e q^2}{m g}} \]
**Assumptions:**
- The group of 27 protons is considered as a point charge situated very closely to one another.
- The lone proton’s interactions are limited to electrostatic
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