/*! 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 33 A small, \(0.60-\mathrm{g}\) bal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 33

A small, \(0.60-\mathrm{g}\) ball in air carries a charge of magnitude \(8.0 \mu \mathrm{C}\). It is suspended by a vertical thread in a downward \(300 \mathrm{~N} / \mathrm{C}\) electric field. What is the tension in the thread if the charge on the ball is \((a)\) positive, (b) negative?

Short Answer

Expert verified
(a) 0.00828 N (b) 0.00828 N

Step by step solution

01

Identify Known Values

The mass of the ball is given as \( m = 0.60 \) grams. To work within the appropriate unit system (SI units), convert the mass to kilograms: \( m = 0.60 / 1000 = 0.0006 \) kg. The charge on the ball is \( q = 8.0 \times 10^{-6} \) C (since \( 1 \mu C = 10^{-6} \) C). The electric field is \( E = 300 \) N/C directed downward.
02

Calculate Gravitational Force

The gravitational force (weight) acting on the ball is calculated using the formula \( F_g = m \cdot g \), where \( g = 9.8 \) m/s\(^2\) is the acceleration due to gravity. Thus, \( F_g = 0.0006 \cdot 9.8 = 0.00588 \) N.
03

Calculate Electric Force for Positive Charge

When the charge is positive, the electric force acting on the ball is in the same direction as the electric field. The electric force is given by \( F_e = q \cdot E = 8.0 \times 10^{-6} \times 300 = 0.0024 \) N downward.
04

Determine Tension in the Thread (Positive Charge)

For equilibrium (the ball is suspended), the tension \( T \) in the thread must balance both the gravitational force and the electric force. Since all forces are downward, \( T = F_g + F_e = 0.00588 + 0.0024 = 0.00828 \) N upward.
05

Calculate Electric Force for Negative Charge

When the charge is negative, the electric force acts opposite to the direction of the electric field, i.e., upward. Therefore, \( F_e = -q \cdot E = -8.0 \times 10^{-6} \times 300 = -0.0024 \) N (upward).
06

Determine Tension in the Thread (Negative Charge)

With the negative charge, the tension must counteract the gravitational force and subtract the upward electric force, giving \( T = F_g - F_e = 0.00588 - (-0.0024) = 0.00828 \) N upward.

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.

Electric Field
An electric field is a region around a charged particle where other charged particles experience a force. It tells us how a charge affects the space around it. The strength of an electric field is measured in newtons per coulomb (N/C).
In this exercise, the electric field is downward at 300 N/C. The ball, carrying a charge, will experience a force due to this field. For a positive charge, the force aligns with the field direction. For a negative charge, the force acts in the opposite direction. This principle is crucial in analyzing the forces on charged particles suspended in electric fields.
Gravitational Force
Gravitational force acts on objects due to Earth鈥檚 gravity. It's calculated as the mass of the object times the gravitational acceleration ( \( g = 9.8 \) m/s\(^2\)). The formula is:
\[ F_g = m \cdot g \]
In the given exercise, the ball's mass is converted to kilograms (0.0006 kg) for proper calculations, leading to a gravitational force \( F_g = 0.00588 \) N. This force always acts downward, pulling objects toward the Earth's center.
Tension Calculation
To find the tension in the thread, we must consider the net forces acting on the ball. Tension is a force within the thread that balances other forces to maintain equilibrium. The calculation depends on whether the charge on the ball is positive or negative.
For a positive charge, both electric and gravitational forces work together downward. Thus, the tension counteracts their sum:
\[ T = F_g + F_e \]
For a negative charge, the electric force opposes the gravitational force. Therefore, the tension is given by the difference:
\[ T = F_g - F_e \]
In both scenarios, the tension is calculated to be 0.00828 N upward, reflecting the force required to keep the ball stationary in the electric field.
SI Units Conversion
Converting units to the International System of Units (SI) ensures consistency and accuracy in calculations. Here, we convert the mass of the ball from grams to kilograms, since SI units use kilograms as the standard for mass.
Mass conversion follows the formula:
\[ m \text{ (kg) } = m \text{ (g) } / 1000 \]
Thus, a 0.60 g mass is equivalent to 0.0006 kg. Similarly, microcoulombs (\( \mu C = 10^{-6} \) C) are converted to coulombs for the charge. Proper unit conversion supports the correct application of formulas and ensures results are dimensionally consistent.

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

Two identical tiny metal balls carry charges of \(+3 \mathrm{nC}\) and \(-12 \mathrm{nC}\). They are \(3 \mathrm{~m}\) apart in vacuum. (a) Compute the force of attraction. ( \(b\) ) The balls are now touched together and then separated to \(3 \mathrm{~cm}\). Describe the forces on them now.

Illustrated in Fig. 24-2, are two identical balls in vaccuum, each of mass \(0.10 \mathrm{~g}\). They carry identical charges and are suspended by two threads of equal length. At equilibrium they position themselves as indicated. Find the charge on either ball. Consider the ball on the left. It is in equilibrium under three forces: (1) the tension \(F_{T}\) in the thread; (2) the force of gravity, $$m g=\left(1.0 \times 10^{-4} \mathrm{~kg}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \times 10^{-4} \mathrm{~N}$$ and (3) the Coulomb repulsion \(F_{E}\). Writing \(\sum F_{x}=0\) and \(\sum F_{y}=0\) for the ball on the left, $$F_{T} \cos 60^{\circ}-F_{E}=0 \quad \text { and } \quad F_{T} \sin 60^{\circ}-m g=0$$ From the second equation, $$F_{T}=\frac{m g}{\sin 60^{\circ}}=\frac{9.8 \times 10^{-4} \mathrm{~N}}{0.866}=1.13 \times 10^{-3} \mathrm{~N}$$ Substituting into the first equation gives $$F_{E}=F_{T} \cos 60^{\circ}=\left(1.13 \times 10^{-3} \mathrm{~N}\right)(0.50)=5.7 \times 10^{-4} \mathrm{~N}$$ But this is the Coulomb force, \(k q q^{\prime} / r^{2}\). Therefore, $$q q^{\prime}=q^{2}=\frac{F_{E} r^{2}}{k}=\frac{\left(5.7 \times 10^{-4} \mathrm{~N}\right)(0.40 \mathrm{~m})^{2}}{9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}$$ from which \(q=0.10 \mu \mathrm{C}\).

Three point charges are placed at the following locations on the \(x\) -axis: \(+2.0 \mu \mathrm{C}\) at \(x=0,-3.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm},-5.0 \mu \mathrm{C}\) at \(x=120 \mathrm{~cm} .\) Find the force \((a)\) on the \(-3.0 \mu \mathrm{C}\) charge,\((b)\) on the \(-5.0 \mu \mathrm{C}\) charge.

Four equal point charges of \(+3.0 \mu \mathrm{C}\) are placed in air at the four corners of a square that is \(40 \mathrm{~cm}\) on a side. Find the force on any one of the charges.

Two small spheres in vacuum are \(1.5 \mathrm{~m}\) apart center-to-center. They carry identical charges. Approximately how large is the charge on each if each sphere experiences a force of \(2 \mathrm{~N}\) ? The diameters of the spheres are small compared to the \(1.5 \mathrm{~m}\) separation. We may therefore approximate them as point charges. Coulomb's Law, \(F_{E}=k_{0} q_{.1} q_{.2} / r^{2}\), leads to $$q_{01} q_{\cdot 2}=q^{2}=\frac{F_{E} r^{2}}{k_{0}}=\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}=5 \times 10^{-10} \mathrm{C}^{2}$$ from which \(q=2 \times 10^{-5} \mathrm{C}\).
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Oscillations

Read Explanation

Thermodynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Rotational 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.