/*! 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 29 (a) What must the charge (sign a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 29

(a) What must the charge (sign and magnitude) of a 1.45 -g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 \(\mathrm{N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Short Answer

Expert verified
(a) The charge is approximately \(2.19 \times 10^{-5} \text{ C}\), positive. (b) The electric field is about \(1.025 \times 10^{-7} \text{ N/C}\).

Step by step solution

01

Understanding Forces

The question involves balancing forces on a charged particle. Since the particle is stationary, the electric force upward equals the gravitational force downward.
02

Calculate Gravitational Force

The gravitational force on the particle can be calculated using \[ F_{gravity} = m \times g \]where \( m = 1.45 \text{ g} = 0.00145 \text{ kg} \) (since 1 g = 0.001 kg) and \( g = 9.8 \text{ m/s}^2 \).Substitute the values to get:\[ F_{gravity} = 0.00145 \times 9.8 = 0.01421 \text{ N} \].
03

Calculate Electric Force

The electric force acting on the charge can be calculated using\[ F_{electric} = q \times E \]Here, \( E = 650 \text{ N/C} \). Since the electric force equals the gravitational force for the particle to remain stationary,\[ q \times 650 = 0.01421 \].
04

Solve for Charge

To find the charge \( q \), rearrange the equation:\[ q = \frac{0.01421}{650} \approx 2.19 \times 10^{-5} \text{ C} \].The charge must be positive, as the electric field is downward, and the charge must exert an upward force to oppose gravity.
05

Calculate Electric Field for Proton

Next, consider the electric field where the electric force on a proton equals its weight. The weight of a proton is \[ F_{gravity} = m \times g \]where \( m = 1.67 \times 10^{-27} \text{ kg} \) is the mass of a proton.\[ F_{gravity} = 1.67 \times 10^{-27} \times 9.8 = 1.64 \times 10^{-26} \text{ N} \].
06

Compute Electric Field Magnitude for Proton

Using the electric force formula\[ F_{electric} = q \times E \],where \( q = 1.6 \times 10^{-19} \text{ C} \) (the charge of a proton), and setting it equal to the gravitational force,\[ 1.6 \times 10^{-19} \times E = 1.64 \times 10^{-26} \],solve for \( E \):\[ E = \frac{1.64 \times 10^{-26}}{1.6 \times 10^{-19}} \approx 1.025 \times 10^{-7} \text{ N/C} \].

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.

Gravitational Force
Gravitational force is an essential concept to understand when analyzing how objects interact with the earth's gravity. All objects with mass experience a gravitational pull exerted by Earth, which pulls them downward. This force is calculated using the formula:
  • Formula for gravitational force: \( F_{gravity} = m \times g \)
  • Where:
    • \( m \) is the mass of the object, measured in kilograms (kg)
    • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) near the Earth's surface
In the exercise, we apply this to a 1.45-g particle, which we convert into kilograms by dividing by 1000. This gives us 0.00145 kg.
This conversion is crucial because the standard unit of mass in physics is the kilogram. By plugging the mass and gravitational acceleration into the formula, we find the gravitational force that needs to be balanced by the electric force to keep the particle stationary.
Electric Field
An electric field is a region around a charged object where another charged object can experience a force. This concept is crucial, especially when dealing with problems involving stationary charges like in the exercise. The strength of the electric field is measured in newtons per coulomb (N/C) and determines the force exerted on the charges within the field.
  • Formula for electric force: \( F_{electric} = q \times E \)
  • Where:
    • \( q \) is the charge of the particle, measured in coulombs (C)
    • \( E \) is the electric field strength
Using the given electric field of 650 N/C, we can calculate the necessary charge for the particle to remain stationary. This is a direct application of how electric fields can counteract gravitational force on a charged particle, showcasing the interesting interplay between different types of forces.
Charge Calculation
Calculating the charge required to counteract gravity involves balancing the electric and gravitational forces acting on a particle. For a particle to remain stationary in an electric field, the electric force must equal the gravitational force acting downward.
The formula used:
  • Formula for electric force: \( F_{electric} = q \times E \)
  • Rearranged to find charge: \( q = \frac{F_{gravity}}{E} \)
In the exercise, we calculated the gravitational force as approximately 0.01421 N.
We set this equal to the product of the charge and the electric field, then solve for the charge \( q \). With an electric field of 650 N/C, our calculation gives the charge as approximately \( 2.19 \times 10^{-5} \text{ C} \). The charge is positive because a downward electric field suggests the need for a charge that produces an upward force, countering the downward pull of gravity. This illustrates how charge calculations can help understand the balance of forces in different settings.

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

cp Strength of the Electric Force. Imagine two 1.0 -g bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electrical repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=\) \(0.600 \mathrm{m}, y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

CP Two positive point charges \(Q\) are held fixed on the \(x\) -axis at \(x=a\) and \(x=-a .\) A third positive point charge \(q,\) with mass \(m,\) is placed on the \(x\) -axis away from the origin at a coordinate \(x\) such that \(|x| << a\) . The charge \(q,\) which is free to move along the \(x\) -axis, is then released. (a) Find the frequency of oscillation of the charge \(q .\) (Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots,\) valid for the case \(|z|<1 .\) (b) Suppose instead that the charge \(q\) were placed on the \(y\) -axis at a coordinate \(y\) such that \(|y| < < a\) , and then released. If this charge is free to move anywhere in the \(x y\) -plane, what will happen to it? Explain your answer.

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{Cis~placed}\) at the point \(x=+0.150 \mathrm{m}, y=0\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m}\)

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Fluids

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

Read Explanation

Medical Physics

Read Explanation

Space Physics

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.