/*! 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 78 Hole in a shell \(*\) Figure \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 78

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.

Short Answer

Expert verified
The electric field at the midpoint of the aperture points outwards of the shell and its magnitude is \(E = \frac{\sigma}{2\epsilon_{0}}\).

Step by step solution

01

Calculating the Field of the Complete Shell

If the missing tiny part were there, the shell would be complete, and its field at any internal point including the point we are interested at, would be 0, as in Gauss's law, the electric field inside a spherical shell is always 0.
02

Calculating the Field created by the Hole's Charge

The field due to the tiny piece could be considered the same as the field of a flat disk of charge, because its radius \(b\) is so much smaller than \(a\). For a disk of charge with a surface charge density \(\sigma\) and radius \(r\), the electric field at a point on the axis of the disk a distance \(d\) away is given by \(E = \frac{\sigma}{2\epsilon_{0}}(1-\frac{d}{\sqrt{d^2+r^2}})\). In this case, \(d=0\), \(r=b\), and the electric field \(E_{disk}\) caused by the charge of the hole at the center of the hole is just \(E_{disk} = \frac{\sigma}{2\epsilon_{0}}\). This field is directed outwards.
03

Superposing the Fields

The field caused by the rest of the shell points inwards and cancels out with the field due to the disk that is going outwards. The electric field at the midpoint of the aperture is the difference between the field due to the disk and the field due to the rest of the shell. So, \(E = E_{disk} - 0 = \frac{\sigma}{2\epsilon_{0}}\).

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 concept in electromagnetism that relates the electric charge within a closed surface to the electric field emanating from it. It helps us calculate the electric field in symmetrical situations.
When a charge is enclosed in a surface, the electric field spreads out uniformly in all directions. For a spherical shell, as in the exercise, the electric field inside such a shell is zero because the contributions from each point on the shell cancel each other out.
This explains why, if the shell were complete without the hole, the electric field at any internal point would be zero. It's crucial to remember this property when dealing with problems involving spherical shells. In summary, Gauss's Law is useful because it allows us to quickly understand complex systems by considering the symmetry and applying simple surface integrals. This can greatly simplify the process of determining electric fields in perfect sphere-like structures.
Superposition Principle
The Superposition Principle is essential when dealing with multiple electric fields in a single scenario. It tells us that the total electric field created by a number of charges is simply the vector sum of the fields created by each charge individually.
In our exercise, we exploit this principle by considering the effect of both the entire shell (if it were complete) and the small disk created by the missing piece.
The total field at the midpoint of the aperture is, therefore, the sum of the field from what remains of the shell and the field due to the missing disk. By examining these individual components separately, and then summing them, we determine the resultant electric field considering both inward and outward contributions.
  • The field from the complete shell would be zero at any internal point.
  • The field from the small missing disk, approximated as lying at the midpoint, is calculated using disk field formulas.
Understanding and applying the Superposition Principle allows us to deconstruct elaborate configurations into simpler, more approachable parts.
Surface Charge Density
Surface Charge Density, represented by the symbol \( \sigma \), is a measure of how much electric charge is distributed over a unit area of a surface. It plays a key role in calculating electric fields, especially in uniformly charged objects like spheres or disks.In the exercise presented, the surface charge density \( \sigma \) aids in determining the effect of both the spherical shell and the small missing piece. This density describes the amount of charge per unit area of the shell.
The Electric Field due to the disk is directly proportional to this surface charge density, as shown in the formula:\[E_{disk} = \frac{\sigma}{2\epsilon_{0}}\]
  • The higher the surface charge density, the stronger the electric field generated by the disk or shell.
  • By understanding \( \sigma \), it becomes straightforward to substitute into electric field equations, estimating how geometric removal or additions impact field strength.
Properly accounting for surface charge density ensures that we grasp how distributions of charge come together to influence resultant fields, like the solution in our exercise.

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

Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

Gauss's law and two point charges ** (a) Two point charges \(q\) are located at positions \(x=\pm \ell\). At points close to the origin on the \(x\) axis, find \(E_{x}\). At points close to the origin on the \(y\) axis, find \(E_{y}\). Make suitable approximations with \(x \ll \ell\) and \(y \ll \ell\) (b) Consider a small cylinder centered at the origin, with its axis along the \(x\) axis. The radius is \(r_{0}\) and the length is \(2 x_{0}\). Using your results from part (a), verify that there is zero flux through the cylinder, as required by Gauss's law.

Zero field in a sphere In Fig. \(1.51\) a sphere with radius \(R\) is centered at the origin, an, infinite cylinder with radius \(R\) has its axis along the \(z\) axis, and an infinite slab with thickness \(2 R\) lies between the planes \(z=-R\) and \(z=R\). The uniform volume densities of these objects are \(\rho_{1}, \rho_{2}\) and \(\rho_{3}\), respectively. The objects are superposed on top of each other; the densities add where the objects overlap. How should the three densities be related so that the electric field is zero everywhere throughout the volume of the sphere? Hint: Find a vector expression for the field inside each object, and then use superposition.

Force between two strips ** (a) The two strips of charge shown in Fig. \(1.47\) have width \(b\), infinite height, and negligible thickness (in the direction perpendicular to the page). Their charge densities per unit area are \(\pm \sigma .\) Find the magnitude of the electric field due to one of the strips, a distance \(x\) away from it (in the plane of the page). (b) Show that the force (per unit height) between the two strips equals \(\sigma^{2} b(\ln 2) / \pi \epsilon_{0}\). Note that this result is finite, even though you will find that the field due to a strip diverges as you get close to it

Field near a stick A stick with length \(2 \ell\) has uniform linear charge density \(\lambda\). Consider a point \(P\), a distance \(\eta \ell\) from the center (where \(0 \leq \eta<1\) ), and an infinitesimal distance away from the stick. Up close, the stick looks infinitely long, as far as the \(\mathrm{E}\) component perpendicular to the stick is concerned. So we have \(E_{\perp}=\lambda / 2 \pi \epsilon_{0} r\). Find the E component parallel to the stick, \(E_{I}\). Does it approach infinity, or does it remain finite at the end of the stick?
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Particle Model of Matter

Read Explanation

Torque and Rotational Motion

Read Explanation

Wave Optics

Read Explanation

Nuclear Physics

Read Explanation

Kinematics Physics

Read Explanation
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.