/*! 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 18 Fields at the surfaces Conside... [FREE SOLUTION] | 91影视

91影视

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.

Short Answer

Expert verified
The electric field on the surface of the sphere is given by \(E_s=\frac{\rho R}{3蔚_0}\), for the cylinder it is \(E_c = \frac{\rho R}{2蔚_0}\), and for the slab, it's \(E_{slab} = \frac{\rho R}{2蔚_0}\). Therefore, \(E_s < E_c = E_{slab}\). The discrepancy happens because the distribution of charges varies based on the configuration of the object.

Step by step solution

01

Electric Field on the Surface of a Sphere

Firstly, we will calculate the electric field on the surface of the sphere. The expression for the electric field \(E_s\) on the surface of a sphere by applying Gauss's Law is: \(E_s = \frac{Q}{4蟺蔚_0 R^2}\). Q is total charge enclosed by the sphere, \(Q = \rho \frac{4}{3}蟺R^3\). Substituting Q in the equation given by Gauss's Law, we get: \(E_s=\frac{\rho R}{3蔚_0}\).
02

Electric Field on the Surface of a Cylinder

For the cylinder, the expression for the electric field \(E_c\) on the surface of the cylinder by applying Gauss's Law is: \(E_c = \frac{位}{2蟺蔚_0 R}\). However, as it is given that the cylinder has an infinite length, the volume becomes area times length, which is denoted as \(位\). Substituting this \(\lambda = \rho \cdot \pi R^2\), we have \(E_c = \frac{\rho R}{2蔚_0}\).
03

Electric Field on the Surface of a Slab

For an infinite slab, we apply Gauss's law on a cylinder formed along the thickness of the slab, where electric field \(E_{slab}\) is: \(E_{slab} = \frac{\sigma}{2蔚_0}\). The Charge per unit area (蟽) for the slab would be given by 蟽 = 蟻R. Hence, \(E_{slab} = \frac{\rho R}{2蔚_0}\).
04

Comparison of the Electric Fields

According to the calculations, it turns out that \(E_c = E_{slab}\) and both are greater than \(E_s\). The reason for this is geometric differences among the three bodies even if they possess the same charge density. For instance, in the sphere, the charges are evenly distributed which reduces the strength of the electric field. However, for the cylinder and the slab, the infinite length contributes to a stronger electric field at their surfaces.

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 91影视!

Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism, stating that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is expressed as \[ \Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \( \Phi_E \) is the electric flux through surface \( S \), \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a vector representing an infinitesimal area on surface \( S \), \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space. In simpler terms, it's a way to calculate the electric field generated by a given charge distribution by using a hypothetical surface known as a Gaussian surface. The choice of Gaussian surface is crucial 鈥 it should be aligned with symmetry to make the calculations easier.

For a sphere, the symmetry leads to a uniform field across the surface, which allows for the aforementioned sphere's electric field calculation. Gauss's Law can also be applied in more complex scenarios like the cylinder and slab in our exercise. The key to using Gauss's Law effectively is to choose an appropriate Gaussian surface that takes advantage of the symmetry of the charge distribution in the problem at hand.
Volume Charge Density
Volume charge density, represented by the symbol \( \rho \), is a measure of how much electric charge is distributed within a given volume. It is defined as the amount of charge per unit volume, or \[ \rho = \frac{Q}{V} \]where \( Q \) is the total electric charge and \( V \) is the volume. In scenarios with uniform charge distribution, volume charge density remains constant across the entire volume.

Understanding volume charge density is crucial when calculating electric fields using Gauss's Law, as it allows us to relate the total charge within a Gaussian surface to its volume. In the exercise provided, all objects 鈥 a sphere, cylinder, and slab 鈥 have the same volume charge density, which indicates that they have the same amount of charge per unit volume. However, due to their different shapes, the distribution of the charge affects the resulting electric field on their surfaces. A key observation in the exercise is that volume charge density alone does not determine the magnitude of the electric field on a surface; the geometry of the charged object must also be taken into account.
Electric Field Calculation
Calculating the electric field involves finding the force that a charge would experience in the space around it. This is typically done by considering the source of the charge and the medium through which the field propagates. For objects with symmetry, such as spheres, cylinders, and slabs, the calculations are simplified by using Gauss's Law.

For the electric field on the sphere's surface, the distribution of charge is symmetric, leading to a simple formula: \[ E_s = \frac{\rho R}{3\varepsilon_0} \]. The cylinder and slab, on the other hand, exhibit different symmetries and, thus, different electric field calculations. The infinite length of the cylinder and the infinite dimensions of the slab alter the impact of the charge's distribution, which is reflected in the formulas \[ E_c = \frac{\rho R}{2\varepsilon_0} \] for the cylinder, and an identical value for the slab.

Why Geometry Affects Electric Field Strength

It's interesting to note that while the volume charge density is the same for all three objects, the electric field values are not. This is due to how the charge is distributed over different geometries. In a sphere, the charge has a spherical distribution, creating a less intense field at the surface. The cylinder and slab, though, with their 'infinite' dimensions, spread the charge over a greater 'surface' at any point, leading to a stronger electric field at their respective surfaces.

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

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

Potential energy of a cylinder A cylindrical volume of radius \(a\) is filled with charge of uniform density \(\rho\). We want to know the potential energy per unit length of this cylinder of charge, that is, the work done per unit length in assembling it. Calculate this by building up the cylinder layer by layer, making use of the fact that the field outside a cylindrical distribution of charge is the same as if all the charge were located on the axis. You will find that the energy per unit length is infinite if the charges are brought in from infinity, so instead assume that they are initially distributed uniformly over a hollow cylinder with large radius \(R\). Write your answer in terms of the charge per unit length of the cylinder, which is \(\lambda=\rho \pi a^{2}\). (See Exercise \(1.83\) for a different method of solving this problem.)

Oscillating on a line ** Two positive point charges \(Q\) are located at points \((\pm \ell, 0) .\) A particle with positive charge \(q\) and mass \(m\) is initially located midway between them and is then given a tiny kick. If it is constrained to move along the line joining the two charges \(Q\), show that it undergoes simple harmonic motion (for small oscillations), and find the frequency.

Oscillating in a ring A ring with radius \(R\) has uniform positive charge density \(\lambda\). A particle with positive charge \(q\) and mass \(m\) is initially located at the center of the ring and is then given a tiny kick. If it is constrained to move in the plane of the ring, show that it undergoes simple harmonic motion (for small oscillations), and find the frequency. Hint: Find the potential energy of the particle when it is at a (small) radius, \(r\), by integrating over the ring, and then take the negative derivative to find the force. You will need to use the law of cosines and also the Taylor series \(1 / \sqrt{1+\epsilon} \approx 1-\epsilon / 2+3 \epsilon^{2} / 8\)

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.