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Two protons placed near one another with no other objects close by would a) accelerate away from each other. d) be pulled together at constant b) remain motionless. speed c) accelerate toward each other. e) move away from each other at constant speed.

Short Answer

Expert verified
a) Accelerate away from each other b) Attract each other c) Remain stationary d) Rotate around each other Answer: a) Accelerate away from each other

Step by step solution

01

Identifying the forces involved

Since protons have positive charges, they interact through electrostatic forces, according to Coulomb's law. The force between two charged particles is given by: F = k * (|q1*q2| / r^2) where F is the electrostatic force, k is Coulomb's constant (approximately 9*10^9 Nm^2/C^2), q1 and q2 are the charges of the two protons, and r is the distance between them.
02

Determine the direction of the force

In this case, both protons have the same charge (positive), so according to Coulomb's law, the electrostatic force between them will be repulsive, meaning they will push each other away.
03

Analyzing the motion of the protons

Since the protons are pushed away from each other by the electrostatic force, they accelerate away from one another. In a vacuum-like environment, no other forces act on the protons, and their acceleration is determined solely by the electrostatic force.
04

Selecting the correct answer

By analyzing the forces involved and their directions, we can determine that when two protons are placed near each other, they will accelerate away from each other. So the correct answer is (a) accelerate away from each other.

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Key Concepts

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

Coulomb's Law
When we explore the interactions between charged particles, one of the fundamental principles we engage with is Coulomb's law. This law quantifies the electrical force between two static, point charges.

It beautifully encapsulates how charged entities behave with an equation: \( F = k \times \frac{\left|q1 \times q2\right|}{r^2} \), where \( F \) is the magnitude of the electrostatic force, \( k \) represents Coulomb's constant (approximately \( 9 \times 10^9 \ Nm^2/C^2 \) in vacuum), \( q1 \) and \( q2 \) are the electric charges of the particles, and \( r \) is the distance between their centers. The forces are attractive if the charges are of opposite sign, and repulsive if the charges are like-sign, as with two protons. The simplicity of this principle makes it vital for understanding the structure of matter and the intricacies of interatomic forces.
Proton Interaction
Looking closely at proton interaction, we know that protons, the positively charged particles in an atomic nucleus, exert an influence on each other. When considering two protons in close proximity, as our problem highlights, they exhibit a mutual repulsion.

This behavior is governed by the electrostatic forces described by Coulomb's law, which predicts that like charges repel. Hence, in the absence of other forces or particles, two protons would accelerate away from each other as they both carry the same type of charge — positive. This interaction is a classic example of electrostatic repulsion, highlighting the key characteristics that play a crucial role in the stability of atomic nuclei and the challenges in nuclear fusion processes.
Electric Charge
To grasp the essence of electric charge, consider it as a fundamental property of particles that determines how they interact with each other and with electromagnetic fields. There are two types of electric charges: positive and negative.

Particles with opposite charges attract each other, whereas particles with like charges repel each other. Protons have a positive charge, which is balanced by the negative charge of electrons in an atom. Charged particles exert electrostatic forces on one another, influencing their motion and position, as described by Coulomb's law. Understanding electric charge is crucial for diving into the world of electromagnetism, and a multitude of technological advancements are based on the principles of charge and its interactions.
Physics Problem Solving
When solving physics problems, a methodical approach is essential for success. By carefully analyzing and breaking down the forces, one can understand complex interactions.

In our exercise, establishing the types of charges and their interaction directs us to the appropriate principles — in this case, Coulomb's law. Recognizing that two protons would repel each other provides clarity on the direction and nature of the force involved.

Next, we analyze the resultant motion, considering only the forces at play. With a clear grasp on these forces, we proceed to articulate an answer that guides us to a deeper comprehension of the physical world. Patients, logical progression, and application of fundamental principles are the cornerstones of effective physics problem solving.

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Most popular questions from this chapter

A particle (charge \(=+19.0 \mu \mathrm{C}\) ) is located on the \(x\) -axis at \(x=-10.0 \mathrm{~cm}\) and a second particle (charge \(=-57.0 \mu \mathrm{C})\) is placed on the \(x\) -axis at \(x=+20.0 \mathrm{~cm} .\) What is the magnitude of the total electrostatic force on a third particle (charge \(=-3.80 \mu \mathrm{C}\) ) placed at the origin \((x=0) ?\)

From collisions with cosmic rays and from the solar wind, the Earth has a net electric charge of approximately \(-6.8 \cdot 10^{5} \mathrm{C}\). Find the charge that must be given to a \(1.0-\mathrm{g}\) object for it to be electrostatically levitated close to the Earth's surface.

Two spheres, each of mass \(M=2.33 \mathrm{~g}\), are attached by pieces of string of length \(L=45.0 \mathrm{~cm}\) to a common point. The strings initially hang straight down, with the spheres just touching one another. An equal amount of charge, \(q\), is placed on each sphere. The resulting forces on the spheres cause each string to hang at an angle of \(\theta=10.0^{\circ}\) from the vertical. Determine \(q,\) the amount of charge on each sphere.

A positive charge \(q_{1}=1.00 \mu \mathrm{C}\) is fixed at the origin, and a second charge \(q_{2}=-2.00 \mu \mathrm{C}\) is fixed at \(x=10.0 \mathrm{~cm} .\) Where along the \(x\) -axis should a third charge be positioned so that it experiences no force?

Two cylindrical glass beads each of mass \(m=10.0 \mathrm{mg}\) are set on their flat ends on a horizontal insulating surface separated by a distance \(d=2.00 \mathrm{~cm} .\) The coefficient of static friction between the beads and the surface is \(\mu_{s}=0.200\). The beads are then given identical charges (magnitude and sign). What is the minimum charge needed to start the beads moving?

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