/*! 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 3 If a proton and an electron are ... [FREE SOLUTION] | 91Ó°ÊÓ

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If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

Short Answer

Expert verified
The electron accelerates at \(6.32 \times 10^{21}\, \text{m/s}^2\) and the proton at \(3.45 \times 10^{18}\, \text{m/s}^2\).

Step by step solution

01

Identify the Forces Acting on Each Particle

The force acting on both the proton and the electron is the electrostatic force. This can be calculated using Coulomb's law, which states that the force between two charged particles is given by \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \), where \( k \) is the electrostatic constant (\( 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the charges of the particles, and \( r \) is the distance between them.
02

Substitute Values into Coulomb's Law

Substitute the known values into the formula: \( q_1 = q_2 = 1.6 \times 10^{-19} \) C (since both the proton and electron have equal but opposite charges) and \( r = 2.0 \times 10^{-10} \) m. The electrostatic force (\( F \)) can be calculated as:\[ F = \frac{(8.9875 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{(2.0 \times 10^{-10})^2} \approx 5.76 \times 10^{-9} \text{ N} \]
03

Calculate the Acceleration of Each Particle

Using Newton's second law \( F = m \cdot a \), we solve for acceleration \( a \) for each particle. Where \( m \) for the electron is \( 9.11 \times 10^{-31} \) kg and for the proton \( 1.67 \times 10^{-27} \) kg.For the electron:\[ a_e = \frac{F}{m_e} = \frac{5.76 \times 10^{-9}}{9.11 \times 10^{-31}} \approx 6.32 \times 10^{21} \text{ m/s}^2 \]For the proton:\[ a_p = \frac{F}{m_p} = \frac{5.76 \times 10^{-9}}{1.67 \times 10^{-27}} \approx 3.45 \times 10^{18} \text{ m/s}^2 \]
04

Confirm the Direction of Acceleration

Since electrostatic forces are attractive between opposite charges, both the proton and the electron will accelerate towards each other.

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Key Concepts

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

Electrostatic Force
The electrostatic force is a fundamental concept when discussing interactions between charged particles. It is the force that arises due to the electric charges of objects, and it can be either attractive or repulsive, depending on the nature of the charges involved.
In our context, we are dealing with a proton and an electron, which carry equal but opposite charges. Therefore, they will experience an attractive force, pulling them towards each other.
Coulomb's law is the essential formula used to calculate the electrostatic force between two point charges. The law is expressed as:
  • \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
where:
  • \( k \) is the electrostatic constant, approximately \( 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \)
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges
  • \( r \) is the distance between the charges
Understanding this force provides critical insight into atomic structures and interactions at a molecular level.
Newton's Second Law
Newton's second law of motion is a cornerstone of classical mechanics. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is concisely expressed as:
  • \( F = m \cdot a \)
where:
  • \( F \) is the net force acting on the object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration of the object
This formula provides a direct way to calculate acceleration when the force and mass are known. In the scenario involving a proton and an electron, despite both particles experiencing the same magnitude of force, their accelerations differ significantly due to their vastly different masses.
Understanding this law allows us to predict how particles will move when subjected to forces.
Acceleration Calculation
Calculating the acceleration of charged particles like a proton and an electron involves applying Newton's second law, as detailed above. Here's how it's done in practice: When these particles are subject to the same electrostatic force, you plug the values into the equation to find acceleration for each.
  • For the electron with a mass of \( 9.11 \times 10^{-31} \text{ kg} \): \[ a_e = \frac{5.76 \times 10^{-9}}{9.11 \times 10^{-31}} = 6.32 \times 10^{21} \text{ m/s}^2 \]
  • For the proton with a mass of \( 1.67 \times 10^{-27} \text{ kg} \):\[ a_p = \frac{5.76 \times 10^{-9}}{1.67 \times 10^{-27}} = 3.45 \times 10^{18} \text{ m/s}^2 \]
These calculations show how dramatically different the accelerations can be.
The electron, being much lighter, accelerates much more rapidly than the heavier proton.
Charged Particles
Understanding the behavior of charged particles such as protons and electrons is vital in physics. These particles form the basic building blocks of all matter.
A charged particle refers to a particle with an electric charge, which can be positive in the case of protons or negative in the case of electrons. The properties of these particles allow atoms to form bonds, enabling the creation of the molecules that compose everything we see.
When charged particles interact, they obey the principles outlined by Coulomb's law and Newton's laws of motion.
Key Considerations:
  • Charged particles generate electric fields
  • Opposite charges attract, similar charges repel
  • The interaction force impacts their motion and can be precisely calculated, as shown in mesh with Newton's laws
Mastering these concepts helps demystify the interactions that govern both micro and macro environments in physics.

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Most popular questions from this chapter

A point charge \(q_1 = -\)4.00 nC is at the point \(x =\) 0.600 m, \(y =\) 0.800 m, and a second point charge \(q_2 = +\)6.00 nC is at the point \(x =\) 0.600 m, \(y =\) 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

In a rectangular coordinate system a positive point charge \(q = 6.00 \times \space 10^{-9} C\) is placed at the point \(x = +0.150 m, y = 0,\) and an identical point charge is placed at \(x = -0.150 m, y = 0.\) 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 m, \(y =\) 0; (c) \(x =\) 0.150 m, \(y = -\)0.400 m; (d) \(x = 0, \)y =$ 0.200 m.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x = 0, \space y = -\)1.35 m; (b) at \(x =\) 12.0 cm, \(y =\) 12.0 cm; (c) at \(x = -\)1.10 m, \(y =\) 2.60 m ? Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).

The earth has a net electric charge that causes a field at points near its surface equal to 150 N\(/\)C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth's electric field a feasible means of flight? Why or why not?

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