/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 \(\bullet\) At what distance wou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 9

\(\bullet\) At what distance would the repulsive force between two electrons have a magnitude of 2.00 \(\mathrm{N}\) ? Between two protons?

Short Answer

Expert verified
The distance for both electrons and protons is the same, approximately 3.86 nm.

Step by step solution

01

Understand Coulomb's Law

Coulomb's Law provides the force between two charges: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( F \) is the force, \( k \) is Coulomb's constant \( (8.988 \times 10^9 \ \mathrm{N \cdot m^2/C^2}) \), \( q_1 \) and \( q_2 \) are the charges (for electrons or protons, this is \( 1.602 \times 10^{-19} \ \mathrm{C} \)), and \( r \) is the distance between the charges.
02

Set Up the Equation for Electrons

Substitute the values into Coulomb's Law to find the distance between two electrons for a force of \( 2.00 \mathrm{N} \); \( F = 2.00 \), \( q_1 = q_2 = 1.602 \times 10^{-19} \). This gives \( 2.00 = 8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{r^2} \).
03

Solve for Distance (Electrons)

Solve for distance \( r \):\[r^2 = 8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{2.00}\]\[r = \sqrt{8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{2.00}}\]Calculate this to find the distance \( r \).
04

Set Up the Equation for Protons

The same setup applies for protons as with electrons: \( q_1 = q_2 = 1.602 \times 10^{-19} \). Solve \( 2.00 = 8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{r^2} \). The process is identical because electrons and protons have the same charge magnitude.
05

Solve for Distance (Protons)

Since the charges are the same, use the same calculation:\[r = \sqrt{8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{2.00}}\]The solved value of \( r \) will be identical to that for electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatics
Electrostatics is a branch of physics that studies stationary or slow-moving electric charges. It focuses on the forces and fields associated with these charges.
  • Two fundamental particles, electrons, and protons, are often studied within this domain due to their opposite charges.
  • Electrons have a negative charge and are found in the space around the nucleus of an atom.
  • Protons have a positive charge and are located within the nucleus of an atom.
  • In electrostatics, like charges repel, and opposite charges attract, resulting in various forces between particles.
The concept of electrostatics allows us to understand the behavior of charged particles, especially when it comes to calculating the forces acting between them.
Force Between Charges
The force between two charges is governed by Coulomb's Law. This fundamental principle tells us how the force works:
  • The formula for the force between two charged particles is given by \[ F = k \frac{|q_1 q_2|}{r^2} \], where:
    • \( F \) is the magnitude of the force.
    • \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
    • \( r \) is the distance between the centers of the two charges.
    • \( k \) is Coulomb's constant, approximately \( 8.988 \times 10^9 \, \mathrm{N \, m^2/C^2} \).
  • This force can be either attractive or repulsive depending on whether the charges are opposite or like.
  • In our example, since both particles are identical in charge (both electrons or both protons), the force is repulsive.
Coulomb's Law is critical for understanding electrostatic interactions at a fundamental level.
Distance Calculation
Calculating the distance between charged particles when a specific force is present involves rearranging Coulomb's Law:
  • Starting with\[ 2.00 = 8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{r^2} \], we need to solve for \( r \).
  • Rearrange the formula to\[r^2 = 8.988 \times 10^9 \cdot \frac{(1.602 \times 10^{-19})^2}{2.00}\]
  • Once you have \( r^2 \), take the square root to find the distance \( r \).
  • For both electrons and protons, the calculation remains the same, as the charge magnitude is identical for both particles.
By calculating \( r \), you can determine how far these particles need to be to exert exactly a \( 2.00 \, \mathrm{N} \) force on each other.
Repulsive Force
A repulsive force occurs when two particles with the same type of charge interact, pushing away from each other.
  • In our problem, both electrons and protons exert repulsive forces because they share identical charges.
  • When applying Coulomb's Law, a positive force value indicates repulsion, consistent with the situation of identical charges.
  • Understanding repulsive forces is essential for explaining why like charges can never stick together naturally.
The concept of repulsive forces underpins much of classical electrostatics, revealing the fascinating interactions shaped by the simple principle that like charges repel.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet$$\bullet\) A point charge of \(-4.00 \mathrm{nC}\) is at the origin, and a second point charge of \(+6.00 \mathrm{nC}\) is on the \(x\) axis at \(x=0.800 \mathrm{m}\) . Find the magnitude and direction of the electric field at each of the following points on the \(x\) axis: (a) \(x=20.0 \mathrm{cm}\) , (b) \(x=1.20 \mathrm{m},(\mathrm{c}) x=-20.0 \mathrm{cm} .\)

\(\bullet\) A negative charge of \(-0.550 \mu C\) exerts an upward 0.200 \(\mathrm{N}\) force on an unknown charge 0.300 \(\mathrm{m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the - 0.550\(\mu \mathrm{C}\) charge?

\(\bullet$$\bullet\) A charge of \(-3.00 \mathrm{nC}\) is placed at the origin of an \(x y-\)coordi- nate system, and a charge of 2.00 \(\mathrm{nC}\) is placed on the \(y\) axis at \(y=4.00 \mathrm{cm} .\) (a) If a third charge, of \(5.00 \mathrm{nC},\) is now placed at the point \(x=3.00 \mathrm{cm}, y=4.00 \mathrm{cm},\) find the \(x\) and \(y\) com- ponents of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

\(\bullet\)(a) How many excess elec- trons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an elec- tric field of 1150 \(\mathrm{N} / \mathrm{C}\) just out- side the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

\(\bullet$$\bullet\) A total charge of magnitude \(Q\) is distributed uniformly within a thick spherical shell of inner radius \(a\) and outer radius b. (a) Use Gauss's law to find the electric field within the cavity \((r \leq a)\) . (b) Use Gauss's law to prove that the electric field outside the shell \((r \geq b)\) is exactly the same as if all the charge were concentrated as a point charge \(Q\) at the center of the sphere. (c) Explain why the result in part (a) for a thick shell is the same as that found in Example 17.10 for a thin shell. A thick shell can be viewed as infinitely many thin shells.)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation

Translational Dynamics

Read Explanation

Engineering Physics

Read Explanation

Magnetism

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.