/*! 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 10 Four electrons are located at th... [FREE SOLUTION] | 91Ó°ÊÓ

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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Short Answer

Expert verified
Work needed is approximately \( 1.03 \times 10^{-18} \, \text{J} \).

Step by step solution

01

Calculate Initial Potential Energy

The initial potential energy of the alpha particle at the center of the square is due to its interaction with the four electrons. The potential energy between the alpha particle and one electron is given by \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) is Coulomb's constant, \( q_1 = +2e \) is the charge of the alpha particle, \( q_2 = -e \) is the charge of an electron, and \( r = \frac{\sqrt{2}a}{2} = 5\sqrt{2} \times 10^{-9} \, \text{m} \) is the distance from the alpha particle to an electron. Calculate \( U \) for each electron and sum the contributions.
02

Calculate Potential Energy at New Position

When the alpha particle is moved to the midpoint of one of the sides of the square, its distance from different electrons changes. Calculate the new distances: two electrons at distance \( a/2 = 5.0 \times 10^{-9} \, \text{m} \), one electron directly opposite the midpoint at distance \( a = 10.0 \times 10^{-9} \, \text{m} \), and one electron at distance \( \sqrt{5}a/2 = 5\sqrt{5} \times 10^{-9} \, \text{m} \). Use the potential energy formula to calculate the potential energy contribution from each electron, and sum these contributions.
03

Calculate Work Done

Work done on the system is the difference in potential energy between the final and initial positions of the alpha particle. Calculate it using:\[ W = U_{final} - U_{initial} \] Substitute the values of final and initial potential energies calculated in steps 1 and 2.

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Key Concepts

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

Coulomb's Law
Coulomb's law is a fundamental principle in electrostatics that helps us understand the forces and energy between charged particles. It describes how the electrostatic force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them. This law is mathematically expressed as:
  • \( F = \frac{k \cdot q_1 \cdot q_2}{r^2} \)
where \( F \) is the force, \( k \) is Coulomb's constant \((8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
Using this law, you understood how to calculate the potential energy between the alpha particle and each of the electrons in the problem. Instead of focusing on the force directly, the exercise makes use of this law to determine the energy states, comparing them to find the work done by moving the alpha particle.
Work-Energy Principle
The work-energy principle is a key concept that connects the work done on an object to its energy. In electrostatics, the work done on a charged particle results in changes in its electrostatic potential energy.
The principle states:
  • The work done on an object is equal to the change in its energy.
  • In our exercise, this translates to: \( W = U_{final} - U_{initial} \).
This formula is essential in finding out the work needed to relocate the alpha particle as described in the exercise.
By calculating the initial potential energy at the center of the square and the potential energy at a new position on the side, the change in energy gives us the work done moving the particle.
Charge Interactions
Charge interactions are central to understanding why particles behave the way they do in an electric field. Like charges repel each other, while opposite charges attract. This fundamental behavior allows us to predict and calculate behavior in systems like the one in the exercise.
  • Electrons, having a negative charge, interact attractively with the positively charged alpha particle.
  • The potential energy calculations made in the exercise depend heavily on these interactions.
By considering how each electron's position changed when the alpha particle is moved, we understand how the potential energy varies.
Charge interactions provide insight into why the alpha particle needs work to be moved to specific positions, as this involves altering the distances over which these attractive forces act.
Alpha Particles
Alpha particles are a type of charged particle consisting of 2 protons and 2 neutrons. They are positively charged due to the protons, making them interact with other charged entities.
  • In this exercise, the alpha particle had a charge of \(+2e\), where \(e\) is the elementary charge.
Understanding the nature of alpha particles helps when applying Coulomb's law to these particles.
Since the alpha particle is positively charged, it is attracted to negatively charged electrons. This interaction defines the potential energy landscape we analyze.

In scenarios involving alpha particles, it is also significant to note their mass is relatively large compared to an electron, affecting how they move under electric forces. However, their charge is what primarily dictates their interaction in electrostatic situations like the one we've explored.

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Most popular questions from this chapter

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

A metal sphere with radius \(R_1\) has a charge \(Q_1\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_2\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A point charge \(q_1 =\) 4.00 nC is placed at the origin, and a second point charge \(q_2 = -\)3.00 nC is placed on the \(x\)-axis at \(x = +\)20.0 cm. A third point charge \(q_3 =\) 2.00 nC is to be placed on the \(x\)-axis between \(q_1\) and \(q_2\) . (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if \(q_3\) is placed at \(x = +\)10.0 cm? (b) Where should \(q_3\) be placed to make the potential energy of the system equal to zero?

A point charge \(q_1 = +\)2.40 \(\mu\)C is held stationary at the origin. A second point charge \(q_2 = -\)4.30 \(\mu\)C moves from the point \(x =\) 0.150 m, \(y =\) 0 to the point \(x =\) 0.250 m, \(y =\) 0.250 m. How much work is done by the electric force on \(q_2\)?

A thin insulating rod is bent into a semicircular arc of radius \(a\), and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

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