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Chapter 18: Problem 13

Two particles, with identical positive charges and a separation of \(2.60 \times 10^{-2} \mathrm{~m}\), are released from rest. Immediately after the release, particle 1 has an acceleration \(\overrightarrow{\mathbf{a}},\) whose magnitude is \(4.60 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), while particle 2 has an acceleration \(\overrightarrow{\mathbf{a}}_{2}\) whose magnitude is \(8.50 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\). Particle 1 has a mass $$ \text { of } 6.00 \times 10^{-6} \mathrm{~kg} . \text { Find } $$ (a) the charge on each particle and (b) the mass of particle 2 .

Short Answer

Expert verified
Charge is \(1.44 \times 10^{-12} \mathrm{C}\); mass of Particle 2 is \(3.25 \times 10^{-6} \mathrm{kg}\).

Step by step solution

01

Identify Key Variables and Constants

First, note the given variables: acceleration of Particle 1 is \(a_1 = 4.60 \times 10^{3} \ \mathrm{m/s}^2\) and its mass is \(m_1 = 6.00 \times 10^{-6} \ \mathrm{kg}\); acceleration of Particle 2 is \(a_2=8.50 \times 10^{3} \ \mathrm{m/s}^2\). Both particles have the same charge \(q\) and are separated by \(r = 2.60 \times 10^{-2} \ \mathrm{m}\). The Coulomb's law constant is \(k = 8.99 \times 10^{9} \ \mathrm{N \cdot m}^2/\mathrm{C}^2\).
02

Apply Newton's Second Law to Particle 1

Using Newton's second law, the force on Particle 1 is \(F_1 = m_1 \cdot a_1\). Calculate the force: \[ F_1 = 6.00 \times 10^{-6} \ \mathrm{kg} \times 4.60 \times 10^{3} \ \mathrm{m/s}^2 = 2.76 \times 10^{-2} \ \mathrm{N} \]
03

Use Coulomb's Law for Particle 1

The electric force acting on Particle 1 due to Particle 2 is given by Coulomb's law: \(F = \frac{k \cdot q^2}{r^2}\). Set \(F = F_1 = 2.76 \times 10^{-2} \ \mathrm{N}\) and solve for \(q^2\): \[ q^2 = \frac{F_1 \cdot r^2}{k} = \frac{2.76 \times 10^{-2} \cdot (2.60 \times 10^{-2})^2}{8.99 \times 10^9} \] \[ q^2 = 2.08 \times 10^{-24} \ \mathrm{C}^2 \] \[ q = \sqrt{2.08 \times 10^{-24}} = 1.44 \times 10^{-12} \ \mathrm{C} \]
04

Apply Newton's Second Law to Particle 2

Using Newton's second law again, the force on Particle 2 is \(F_2 = m_2 \cdot a_2\). Since the forces between the particles are equal, \(F_2 = F_1 = 2.76 \times 10^{-2} \ \mathrm{N}\). We can set this equal to \(m_2 \cdot a_2\) and solve for \(m_2\): \[ m_2 = \frac{F_2}{a_2} = \frac{2.76 \times 10^{-2}}{8.50 \times 10^{3}} = 3.25 \times 10^{-6} \ \mathrm{kg} \]
05

Conclude the Findings

Each particle has a charge of \(1.44 \times 10^{-12} \ \mathrm{C}\), and the mass of particle 2 is \(3.25 \times 10^{-6} \ \mathrm{kg}\).

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Key Concepts

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

Electric Force
The electric force is a fundamental interaction between any two charged particles. This force can be attractive or repulsive depending on the charges involved. Coulomb's Law describes the magnitude of the electric force between two point charges. It states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them. In formulaic terms, it is expressed as:\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]where:- \( F \) is the magnitude of the force between the charges,- \( q_1 \) and \( q_2 \) are the charges,- \( r \) is the separation between the charges,- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2} \).In our problem, the force calculated affects both particles equally due to their identical charges. The knowledge of the electric force helps us find the acceleration and further properties of the particles.
Newton's Second Law
Newton's second law of motion is a principle that states the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This relation is given by the equation:\[ F = m \cdot a \]where:- \( F \) represents the net force acting on the object,- \( m \) is the mass of the object,- \( a \) is the acceleration of the object.This law is crucial when analyzing particle dynamics under the influence of force. In the exercise at hand, Newton's second law enables us to calculate the force experienced by each particle using their respective masses and accelerations. By equating this with the electric force from Coulomb鈥檚 law, we can extract further information such as the charges and mass differences between the particles.
Particle Acceleration
When a force acts on a particle, it causes the particle to accelerate. The amount of acceleration depends on the force's magnitude and the particle's mass, as explained by Newton's second law. In this particular problem, one particle accelerates at a rate of \( 4.60 \times 10^3 \ \mathrm{m/s^2} \), while the other accelerates at \( 8.50 \times 10^3 \ \mathrm{m/s^2} \). This difference in acceleration, despite having the same charge, is due to a difference in their masses.
  • Particle 1 is less massive, thus accelerates less compared to Particle 2 under the same force.
  • Knowing each particle's acceleration allows us to calculate each particle's individual force, showing how electrical interactions can drive motion.
This concept is important as it connects electrical interactions to kinematic quantities like acceleration.
Charge Calculation
The calculation of the charge follows from rearranging Coulomb's law so that the electric force equals the force calculated via Newton's second law. In our exercise, it involves:1. Setting up the equality from both laws, substituting known values.2. Solving for \( q^2 \), then obtaining \( q \) by taking the square root.Given the electric force experienced is \( 2.76 \times 10^{-2} \mathrm{N} \), and using the separation and Coulomb's constant in the equation:\[ q^2 = \frac{F_1 \cdot r^2}{k} = \frac{2.76 \times 10^{-2} \cdot (2.60 \times 10^{-2})^2}{8.99 \times 10^9} \]\[ q = \sqrt{2.08 \times 10^{-24}} = 1.44 \times 10^{-12} \ \mathrm{C} \]This calculation tells us the amount of charge on each particle, crucial for predicting and understanding their behavior in other potential electric fields.

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Most popular questions from this chapter

Two particles are in a uniform electric field whose value is \(+2500 \mathrm{~N} / \mathrm{C}\). The mass and charge of particle 1 are \(m_{1}=1.4 \times 10^{-5} \mathrm{~kg}\) and \(q_{1}=-7.0 \mu \mathrm{C},\) while the corresponding values for particle 2 are \(m_{2}=2.6 \times 10^{-5} \mathrm{~kg}\) and \(q_{2}=+18 \mu \mathrm{C} .\) Initially the particles are at rest. The particles are both located on the same electric field line, but are separated from each other by a distance \(d\). When released, they accelerate, but always remain at this same distance from each other. Find \(d\).

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. How would you determine the magnitude and direction of the electric field if the object were (a) positively charged and (b) negatively charged?

Interactive Solution \(18.37\) at provides a model for problems of this kind. A small object has a mass of \(3.0 \times 10^{-3} \mathrm{~kg}\) and a charge of \(-34 \mu \mathrm{C}\). It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.

Two tiny conducting spheres are identical and carry charges of \(-20.0 \mu \mathrm{C}\) and \(+50.0 \mu \mathrm{C}\). They are separated by a distance of \(2.50 \mathrm{~cm}\). (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of \(2.50 \mathrm{~cm}\). Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear momentum to \(5.0 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) from \(1.5 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) in a time of \(6.3 \times 10^{-6} \mathrm{~s}\). What is the magnitude of the electric field?
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