/*! 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 \(\mathrm{nm}\) 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
The work needed is \(-4.6 \times 10^{-19} \text{ J}\).

Step by step solution

01

Identify Charges and Configuration

We have four electrons at the corners of a square, each with a charge of \( -e \), where \( e = 1.602 \times 10^{-19} \text{ C} \). An alpha particle with a charge of \( 2e \) is at the center. The side length of the square is 10 \( \text{nm} \), and we want to move the alpha particle from the center to the midpoint of one side.
02

Calculate Initial Potential Energy

The potential energy \( U \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb's constant. Since all four electrons interact with the alpha particle, calculate the distance from the center to a corner: \( a = \frac{10}{2} \sqrt{2} = 5\sqrt{2} \text{ nm} \).
03

Compute Total Initial Potential Energy

Using the formula for potential energy, calculate the initial energy at the center. Each electron-\( \alpha \) interaction contributes: \[ U = 4 \cdot \frac{k \cdot (-e) \cdot 2e}{a} \]Where \( a = 5 \sqrt{2} \text{ nm} = 5\sqrt{2} \times 10^{-9} \text{ m} \). Initialize the calculation to find \( U_{ ext{initial}} \).
04

Calculate Final Potential Energy

Move the \( \alpha \) particle to the midpoint of a side. The distance from this point to the two nearest electrons is \( 5 \text{ nm} \) and to the diagonal ones: \[ a_f = \sqrt{(5^2 + 10^2)} \text{ nm} = \sqrt{125} = 5\sqrt{5} \text{ nm} \].Calculate potential energy at this new position: \[ U_{ ext{final}} = 2 \cdot \frac{k \cdot (-e) \cdot 2e}{5 \times 10^{-9}} + 2 \cdot \frac{k \cdot (-e) \cdot 2e}{5\sqrt{5} \times 10^{-9}} \]
05

Determine Work Done

The work done \( W \) is the change in potential energy:\[ W = U_{ ext{final}} - U_{ ext{initial}} \]Substitute the calculated values to find the numerical result for \( W \). Evaluate to finalize the amount of work needed.

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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 describing the force between two point charges. When you're dealing with charged particles, this law becomes crucial.
The force between any two charges can be calculated using the formula:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:
  • \(F\) is the Coulombic force.
  • \(k\) is the Coulomb's constant, \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
  • \(|q_1\cdot q_2|\) are the magnitudes of the charges.
  • \(r\) is the distance between the charges.
This law helps determine the attractions or repulsions between charges. In the given exercise, it aids in understanding how an alpha particle interacts with electrons at the corners of a square.
Work Done in Electric Fields
When a charge moves within an electric field, work is involved. Work (\(W\)) in the electric field context is the energy expended to move a charge from one point to another.
The work done can be expressed as the change in electrostatic potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]Here:
  • \(U_{\text{final}}\) is the potential energy at the ending position.
  • \(U_{\text{initial}}\) is the potential energy at the starting position.
In our scenario, calculating the work involves determining how much energy is needed to shift the alpha particle from the center to the midpoint of a square's side.
By computing the initial and final potential energies, you can find how much effort, or work, the process requires.
Alpha Particle Movement
Alpha particles, which consist of 2 protons and 2 neutrons, have a charge of \(+2e\). They play a central role in this exercise due to their movement from one part of the electric field to another.
Understanding alpha particle movement is essential as it shows how charged particles interact under forces like those defined by Coulomb's Law.
In this problem, moving the alpha particle from the square's center where it starts to the midpoint of a side means changing its position within the configuration of electrons. This shift affects the potential energy, highlighting the principles of energy conservation and forces in electric fields.
Charge Configuration
Charge configuration refers to the arrangement of charged particles in space. It's essential for determining how particles interact.
In exercises like this one:
  • The configuration is initially given by four electrons at the square's corners.
  • An alpha particle sits at the center.
  • Electrons, being negatively charged, create a force field affecting the positively charged alpha particle.
Recognizing this spatial setup allows one to calculate the distances and interactions between each charge, necessary for finding potential energy and work done. The symmetric arrangement simplifies calculations while understanding how each electron affects the alpha particle's movement.
Potential Energy Calculation
Calculating the potential energy between charged particles provides insight into their interactions.
Potential energy \(U\) between two point charges is expressed by:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where:
  • \(k\) is Coulomb's constant.
  • \(q_1\) and \(q_2\) are the charges involved.
  • \(r\) is the separation distance between them.
Using this formula, the exercise requires computing the potential energy at two positions — the center of the square and midway along one side.
Summing individual interactions between the alpha particle and each electron, both initially and after movement, you determine the total initial and final potentials. The difference between these quantities ultimately tells you how much work is required to move the alpha particle, crucial for understanding energy changes in electric fields.

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Most popular questions from this chapter

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 \(\mathrm{m} / \mathrm{s} ?\)

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

Determining the Size of the Nucless. When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons. (a) What was the electric potential energy of the alpha-radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.

Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are 0.500 \(\mathrm{m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d .\) Two of the point charges are identical and have charge \(q .\) If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

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