/*! 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 Two small plastic spheres are gi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 9

Two small plastic spheres are given positive electric charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Short Answer

Expert verified
(a) Each sphere has a charge of approximately \( 7.42 \times 10^{-6} \text{ C} \). (b) The smaller charge is \( 3.71 \times 10^{-6} \text{ C} \), and the larger charge is \( 1.48 \times 10^{-5} \text{ C} \).

Step by step solution

01

Understanding Coulomb's Law

The force exerted between two point charges is defined by Coulomb's law, given by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the force between the charges, \( k \) is Coulomb's constant \((8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
02

Calculation for Equal Charges \( (q_1 = q_2 = q) \)

If the charges are equal, \( q_1 = q_2 = q \). Plug \( q_1 = q_2 = q \) and \( r = 0.15 \text{ m} \) into the formula: \[ 0.220 = (8.988 \times 10^9) \frac{q^2}{(0.15)^2} \]. Solving for \( q \), we find \[ q^2 = \frac{0.220 \times (0.15)^2}{8.988 \times 10^9} \]. Compute \( q \) by taking the square root of the right side of the equation.
03

Solving for \( q \): Equal Charges

Calculate \( q^2 \) which equals approximately \( 5.50 \times 10^{-12} \). Taking the square root, we find \( q \approx 7.42 \times 10^{-6} \text{ C} \). So each sphere has a charge of approximately \( 7.42 \times 10^{-6} \text{ C} \).
04

Understanding Unequal Charge Distribution

For case (b), where one charge is four times the other, denote \( q_1 = q \) and \( q_2 = 4q \). Substitute into the formula: \[ 0.220 = (8.988 \times 10^9) \frac{q \cdot 4q}{(0.15)^2} \].
05

Solving for \( q \): One Charge is Four Times the Other

The expression becomes \[ 0.220 = (8.988 \times 10^9) \frac{4q^2}{(0.15)^2} \]. Solving for \( q^2 \), we get \[ q^2 = \frac{0.220 \times (0.15)^2}{4 \times 8.988 \times 10^9} \]. Calculate \( q^2 \) and find \( q \approx 3.71 \times 10^{-6} \text{ C} \). Thus, \( q_1 = 3.71 \times 10^{-6} \text{ C} \) and \( q_2 = 1.48 \times 10^{-5} \text{ C} \).

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.

Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field.
It is measured in coulombs (C) and two types exist: positive and negative.
This property is pivotal in determining how objects will interact through electric forces.
  • Like charges, both positive or both negative, will repel each other.
  • Opposite charges will attract each other, such as a positive and a negative charge.
The amount of charge on an object can vary, influencing the magnitude of forces it experiences and exerts.
For example, in the exercise, two plastic spheres with positive charges repel each other. Knowing their charges helps us understand and calculate the force experienced.
Repulsive Force
The repulsive force is observed between like charges, where two similarly charged objects push away from each other. This force can be measured and calculated using Coulomb's Law, as seen in the given exercise.
This type of force is opposite to an attractive force, which occurs between unlike charges.
  • Repulsion occurs when particles have charges of the same sign.
  • The size of the repulsive force increases with greater charge magnitudes and decreases with the square of the distance between them.
This principle is crucial in various applications, including electrostatics and electricity. Understanding the nature of these forces allows us to manipulate electronic devices and understand natural phenomena.
Point Charges
Point charges are idealized charges located at a single point in space, making the analysis of electric forces much simpler.
They are used in theoretical and instructional contexts to focus on the principles of charge and force without the complexities of real-world charge distributions.
  • They are useful for calculating forces and electric fields in controlled scenarios.
  • The concept assumes all charge is concentrated at a single point with no physical size.
In the exercise, two charged spheres are treated as point charges to apply Coulomb's Law effectively.
This allows us to easily calculate the forces between them based on their charge values and separation distance.
Coulomb's Constant
Coulomb's constant, denoted as \( k \), is a proportionality factor in Coulomb's Law.
It relates the force between two point charges to the product of their charges and the inverse square of the distance between them.
Its value is approximately \( 8.988 \times 10^9 \) Nm虏/C虏.
  • It provides a means to calculate the strength of the electric force in the International System of Units (SI).
  • The constant simplifies the calculation of electric forces in theoretical problems and practical applications.
By understanding Coulomb's constant, students can more accurately predict and quantify electric interactions, as demonstrated in the step-by-step solution to the given exercise.

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

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_0\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{max}\) that the proton descends vertically below its initial elevation. Ignore gravitational forces. (b) After what horizontal distance d does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{max}\) and \(d\) if \(E =\) 500 N\(/\)C, \(v_0 =\) 4.00 \(\times 10^5\) m\(/\)s, and \(\alpha =\) 30.0\(^\circ\).

A \(-\)3.00-nC point charge is on the \(x\)-axis at \(x =\) 1.20 m. A second point charge, \(Q,\) is on the \(x\)-axis at -0.600 m. What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 N\(/\)C in the \(+\)x-direction, (b) 45.0 N\(/\)C in the \(-\)x-direction?

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 \(\mu\)s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 mm away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Turning Points in Physics

Read Explanation

Modern Physics

Read Explanation

Translational 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.