/*! 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 27 A charge of \(-3.00 \mu \mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 27

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

Short Answer

Expert verified
The particle travels 0.0308 meters before stopping.

Step by step solution

01

Identify Forces and Energy

This scenario involves electric forces and principles of energy conservation. Initially, the particle has kinetic energy due to its speed, and as it moves towards the fixed charge, it experiences electric repulsion because both charges are negative.
02

Write Energy Conservation Equation

The initial kinetic energy, \( KE_i = \frac{1}{2} m v^2 \), is converted into electric potential energy, \( U = \frac{k |q_1 q_2|}{r} \), where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them. The equation is: \( \frac{1}{2} m v^2 = \frac{k |q_1 q_2|}{r} \).
03

Insert Known Values

\( m = 7.20 \times 10^{-3} \text{ kg} \), \( v = 65.0 \text{ m/s} \), \( q_1 = 3.00 \times 10^{-6} \text{ C} \), \( q_2 = 8.00 \times 10^{-6} \text{ C} \), and \( k = 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \). Plug these into the equation: \( \frac{1}{2} \times 7.20 \times 10^{-3} \times 65^2 = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-6} \times 8.00 \times 10^{-6}}{r} \).
04

Solve for Distance \( r \)

Calculate the left side: \( \frac{1}{2} \times 7.20 \times 10^{-3} \times 4225 = 15.21 \text{ J} \). Calculate the right side: \( 8.99 \times 10^9 \times 24 \times 10^{-12} = 215.76 \times 10^{-3} \). Now solve for \( r \): \( 15.21 = \frac{215.76 \times 10^{-3}}{r} \). Rearranging gives \( r = \frac{215.76 \times 10^{-3}}{15.21} \approx 0.0142 \text{ m} \).
05

Calculate Total Distance Traveled

The distance traveled before the speed is zero is initially 0.0450 m. The particle stops at 0.0450 - 0.0142. Calculate: \( 0.0450 - 0.0142 = 0.0308 \text{ m} \).

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.

Energy Conservation
Energy conservation is a principle stating that the total energy in a closed system remains constant over time. In this exercise, the principle is used to understand how energy transforms from one form to another.
Initially, the particle has kinetic energy created by its motion. As it moves towards the fixed charge, it feels an electric repulsion because of the similar negative charges.
- **Kinetic Energy to Potential Energy conversion:** The initial kinetic energy decreases as some of it is transformed into electric potential energy.
- **Energy Balance:** At the point where the particle's speed becomes zero, all kinetic energy has been converted into electric potential energy.
This principle helps find the point where the particle stops, using an energy conservation equation that balances kinetic and electric potential energies.
Kinetic Energy
Kinetic energy is the energy that an object possesses because of its motion. For any moving object, this energy is calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where **m** is the mass of the object and **v** is its velocity.
  • As the particle in this exercise is shot with an initial speed of 65 m/s, it has a significant amount of kinetic energy initially.
  • The mass of the particle, given as \(7.20 \times 10^{-3}\) kg, contrasts with its high velocity to determine its initial kinetic energy value.
This initial kinetic energy is an essential part of understanding how the particle's motion will evolve as it approaches the fixed charge.
Electric Potential Energy
Electric potential energy is the energy stored in charged particles due to their positions within an electric field. In this context, it arises because of the interaction between two charges.
When two charges are brought near each other, like in the exercise:
  • There is an electric force of repulsion because both charges are negative.
  • The potential energy can be calculated using \( U = \frac{k |q_1 q_2|}{r} \).
- **Formula components:** - **k**: Coulomb鈥檚 constant, \(8.99 \times 10^9\) N m虏/C虏, - **q鈧�** and **q鈧�**: The magnitudes of the charges (\(-3.00\) 渭C and \(-8.00\) 渭C, respectively), and - **r**: The separation distance between the charges.As the particle moves towards the fixed charge, its speed decreases due to an increase in potential energy, eventually bringing it to a stop when all kinetic energy is converted into potential energy.

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

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text {cloud }}-V_{\text {ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

At a distance of \(1.60 \mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there are additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V} .\) Starting at a distance of \(1.60 \mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include thenstarting surface.

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m} .\) What is the magnitude of the electric field in the membrane?

The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store \(11 \mu \mathrm{C}\) of charge and \(5.0 \times 10^{-5} \mathrm{J}\) of energy. When used with capacitor \(\mathrm{B}\) which has a capacitance of \(6.7 \mu \mathrm{F}\), this voltage causes the capacitor to store a charge that has a magnitude of \(q_{\mathrm{B}} .\) Determine \(q_{\mathrm{B}}\).

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J},\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J} .\) The voltage across the plates of capacitor B is 12 V. Find the voltage across the plates of capacitor A.
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Translational Dynamics

Read Explanation

Waves Physics

Read Explanation

Physics of Motion

Read Explanation

Dynamics

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.