/*! 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 14 22\. 14. Electric Fields in an A... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

22\. 14. Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?(\mathrm{c})\) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

Short Answer

Expert verified
The electric field at the nucleus is zero due to the electrons, strong just outside the nucleus, and weaker at the electrons' distance.

Step by step solution

01

Calculate the Charge of the Nucleus

First, calculate the total charge of the uranium nucleus. A uranium nucleus contains 92 protons, and each proton has a charge of \( e = 1.602 \times 10^{-19} \text{C} \). The total charge \( Q \) is given by \( Q = 92 \times e = 92 \times 1.602 \times 10^{-19} \text{C} \).
02

Determine the Electric Field Just Outside the Nucleus

The electric field just outside the surface of a spherically symmetric charged sphere is given by the formula: \[ E = \frac{1}{4\pi\varepsilon_0} \times \frac{Q}{r^2} \]where \( r = 7.4 \times 10^{-15} \text{ m} \) is the radius of the nucleus, and \( \varepsilon_0 \) is the vacuum permittivity \( = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \). Substitute the known values to find \( E \).
03

Calculate the Field at Electron Distance

Use the same formula for the electric field with a different \( r \) to find the electric field at the distance of the electrons: \[ E = \frac{1}{4\pi\varepsilon_0} \times \frac{Q}{r^2} \]where \( r = 1.0 \times 10^{-10} \text{ m} \). Using the previously calculated \( Q \), substitute \( r \) with the new distance.
04

Calculate the Net Electric Field at the Nucleus Due to Electrons

The electrons surrounding a nucleus can be approximated as a uniform shell of charge. By Gauss's law, a charge shell produces no electric field inside itself. Therefore, at the location of the nucleus, the electric field due to the electrons is zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Uranium Nucleus
The uranium nucleus is a fascinating structure in the field of nuclear physics. It is composed of protons and neutrons and houses 92 protons. Each proton carries a positive charge of approximately \(1.602 \times 10^{-19} \text{ C}\). Thus, the total charge of the uranium nucleus can be calculated by multiplying this charge with the number of protons, leading to a significant positive charge.
The nucleus is extremely small, with a radius of about \(7.4 \times 10^{-15} \text{ m}\). Despite its tiny size, it is packed with energy, due to both the mass of the protons and neutrons, and the electric potential energy generated by the charges.Being a primary source of charge, a uranium nucleus generates an electric field just outside its surface. Understanding this field is essential for grasping the interactions within an atom and the forces that hold the nucleus together.
Spherically Symmetric Charge
When we say the uranium nucleus is a "spherically symmetric charge," we imply that the charge density is evenly distributed across the spherical volume of the nucleus. This symmetry simplifies the calculations of electric fields around it.
By considering the charge as spherically symmetric, we use the formula for scalar electric fields, avoiding complex vector calculations. This symmetry allows us to treat the nucleus as if all of its charge were concentrated at its center when determining the electric field at points outside the nucleus.
  • The symmetry results in identical electric field magnitudes at any point equidistant from the nucleus.
  • This uniformity implies a predictable electric potential energy surrounding the nucleus, essential for understanding atomic behavior and reactions.
Such considerations help predict forces and energies within atoms, crucial for imagining nuclear reactions and bonding scenarios in chemistry and physics.
Gauss's Law
Gauss's law is a pivotal concept when dealing with electric fields emanating from symmetric charge distributions, like the uranium nucleus. It states that the net electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is expressed as: \[ \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \(\Phi\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a differential area of the closed surface, \(Q_{\text{enc}}\) is the total charge enclosed within the surface, and \(\varepsilon_0\) is the vacuum permittivity.
In the case of the uranium nucleus, Gauss’s law helps to confirm that a spherically symmetric charge produces an electric field that remains unchanged within a hollow symmetric shell. This means that if you are inside a uniformly charged spherical shell, the net electric field felt is zero, highlighting the nature of electric shielding.
Vacuum Permittivity
Vacuum permittivity, symbolized as \(\varepsilon_0\), is a fundamental physical constant crucial in the computation of electric fields. It characterizes the capacity of the vacuum to permit electric field lines. The value of vacuum permittivity is approximately \(8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\).
This constant appears in Gauss's law and Coulomb's law, linking electric field strength to the charge that generates it. It provides a scale to measure how much electric field can pass through a medium, with vacuum being the least resistive medium possible.
Understanding vacuum permittivity is essential for grasping electric interactions in a vacuum. It frames the scale and strength of interactions between charged particles, making it a vital element in fields like electromagnetism, atomic physics, and electrical engineering. By standardizing the interaction magnitude, \(\varepsilon_0\) allows consistent calculation and comparison of electric fields across different scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

22.9. A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{cm},\) giving it a charge of \(-15.0 \mu \mathrm{C}\) . Find the electric field (a) just inside the paint layer; (b) just outside the paint layer: (c) 5.00 \(\mathrm{cm}\) outside the surface of the paint layer.

22.24. A point charge of \(-2.00 \mu \mathrm{C}\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3} .\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

22.10. A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m}\) . What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radins (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

22.54. A Uniformiy Charged Stab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) - and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho\) . (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

22.50. (a) How many excess electrons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface of the sphere? (b) What is the electric field at a point 10.0 \(\mathrm{cm}\) outside the surface of the sphere?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.