/*! 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 25 Two equal fields * The result ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two equal fields * The result of Exercise \(1.78\) is that the electric field at the center of a small hole in a spherical shell equals \(\sigma / 2 \epsilon_{0}\). This happens to be the same as the field due to an infinite flat sheet with the same density \(\sigma\). That is, at the center of the hole at the top of the spherical shell in Fig. 1.39, the field from the shell equals the field from the infinite horizontal sheet shown. (This sheet could actually be located at any height.) Demonstrate this equality by explaining why the rings on the shell and sheet that are associated with the angle \(\theta\) and angular width \(d \theta\) yield the same field at the top of the shell.

Short Answer

Expert verified
The fields equate due to the fact that for every ring with an angular width of \( d \theta \) on the spherical shell, there is a corresponding ring in a matching position on the infinite flat sheet. The fields due to each contribute equally to the total field because the distances, charges, and angles associated with each ring correspond. Consequently, the fields from an infinite flat sheet and the shell at the center of the hole are equal.

Step by step solution

01

Define the problem and relevant variables

First, let's specify the problem. We have an infinite flat sheet with a charge density \( \sigma \). And we have a spherical shell with a small hole at the center. The goal is to prove that the field at the center of the hole is equal to the field due to the infinite flat sheet.
02

Calculate the field due to the spherical shell

We'll start with the known field due to the shell. As we know from Exercise 1.78, the electric field \( E \) at the center of a small hole of a charged spherical shell is \( \sigma / 2 \epsilon_{0} \) where \( \sigma \) is the surface charge density
03

Calculate the field due to the infinite flat sheet

Next, let's calculate the field due to the infinite flat sheet. The field \( E \) due to an infinite flat sheet of charge is also \( \sigma / 2 \epsilon_{0} \). This equation derives from the superposition of contributions to the field from each small amount of charge \( dQ \) of the sheet.
04

Ultimately the fields equate

For every ring with an angular width of \( d \theta \) on the shell, there is a corresponding ring on the infinite flat sheet. Because the distances, charges, and angles all correspond, the fields due to each contribute equally, ultimately making the fields from an infinite flat sheet and the shell equal at the center of the hole.

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Key Concepts

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

Spherical Shell
A spherical shell is a three-dimensional, hollow object that surrounds a central point and has a symmetrical, spherical shape. Imagine it like a thin crust of a sphere that contains its charge on the surface. In the context of electric fields, a charged spherical shell can be used to study how charge is distributed over surfaces and how these charges affect points in and around the sphere.
  • A spherical shell might have a uniform surface charge distribution, where charge is spread evenly over its surface.
  • When considering electric fields produced inside and outside a spherical shell, it's important to remember shell symmetry—this greatly simplifies calculations.
Gauss's Law is particularly useful with spherical shells because the symmetry allows for simplification. You can determine the electric field inside, outside, and on the surface with ease steps.
Charge Density
Charge density is a measure of how much electric charge is present per unit area on a surface or per unit volume in a space. In equations, it can be described using the Greek letter sigma (\(\sigma \)) for surface charge densities.
  • Surface charge density, \(\sigma \), refers to how charge is distributed over a two-dimensional surface, like a sheet or a spherical shell.
  • Volume charge density extends this idea to three dimensions, indicating charge per unit volume but is not directly relevant to our scenario.
When dealing with charge densities on surfaces like spherical shells or sheets, the uniformity of the \(\sigma \) matters greatly. Uniform charge densities allow us to use symmetric arguments and calculate electric fields using whole regions instead of summing up individual point charges.
Gauss's Law
Gauss's Law is a key principle in electromagnetism that connects electric fields and charge distributions. It states that the electric flux through a closed surface is proportional to the enclosed electric charge. Mathematically, Gauss's Law is expressed as:
\[\oint E \cdot dA = \frac{Q_{enc}}{\epsilon_0}\]where \(E\) is the electric field, \(dA\) is an elemental area vector on the closed surface, \(Q_{enc}\) is the total charge enclosed, and\(\epsilon_0 \) is the permittivity of free space.
  • Gauss's Law is primarily useful when a problem has high symmetry, such as spherical or cylindrical symmetry.
  • In scenarios like spherical shells with symmetry, Gauss's Law simplifies the computation of electric fields.
In our example of a spherical shell with a hole, symmetry allows us to elegantly demonstrate that the electric field results from similar charge distributions, even when comparing to a flat, infinite sheet.
Infinite Plane
An infinite plane in electrostatics is an idealized, flat, two-dimensional surface that extends indefinitely in all directions. This idea allows for simplifications in calculating electric fields due to symmetric properties.
  • The infinite nature simplifies calculations because at any point above the plane, the field can be considered uniform.
  • For an infinite plane with charge density \(\sigma \), the electric field it produces is always \(\sigma / 2\epsilon_0\), directed perpendicularly away from the plane.
The beauty of using this concept is the simplification of the electric field calculations across various systems. By considering a similarly charged spherical shell and infinite plane, this uniformity results in identical fields at certain symmetrical points, like the center of a small hole in our spherical shell exercise.

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Most popular questions from this chapter

Field from a hemisphere ** (a) What is the electric field at the center of a hollow hemispherical shell with radius \(R\) and uniform surface charge density \(\sigma\) ? (This is a special case of Problem \(1.12\), but you can solve the present exercise much more easily from scratch, without going through all the messy integrals of Problem 1.12.) (b) Use your result to show that the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\) equals \(\rho R / 4 \epsilon_{0}\)

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.

(a) Two rings with radius \(r\) have charge \(Q\) and \(-Q\) uniformly distributed around them. The rings are parallel and located a distance \(h\) apart, as shown in Fig. \(1.35\). Let \(z\) be the vertical coordinate, with \(z=0\) taken to be at the center of the lower ring. As a function of \(z\), what is the electric field at points on the axis of the rings? (b) You should find that the electric field is an even function with respect to the \(z=h / 2\) point midway between the rings. This implies that, at this point, the field has a local extremum as a function of \(z\). The field is therefore fairly uniform there; there are no variations to first order in the distance along the axis from the midpoint. What should \(r\) be in terms of \(h\) so that the field is very uniform? By "very" uniform we mean that additionally there aren't any variations to second order in \(z\). That is, the second derivative vanishes. This then implies that the leading-order change is fourth order in \(z\) (because there are no variations at any odd order, since the field is an even function around the midpoint). Feel free to calculate the derivatives with a computer.

Building a sheet from rods ** An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\), integrate over these rods to show that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\)

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