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Chapter 22: Problem 53

An insulating hollow sphere has inner radius \(a\) and outer radius \(b .\) Within the insulating material the volume charge density is given by \(\rho(r)=\frac{\alpha}{r},\) where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a,\) what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a < r < b ?\) (b) A point charge \(q\) is placed at the center of the hollow space, at \(r=0 .\) In terms of \(\alpha\) and \(a,\) what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a < r < b,\) and what then is the value of the constant field in this region?

Short Answer

Expert verified
(a) \( E = \frac{\alpha(r^2 - a^2)}{2\varepsilon_0 r^2} \). (b) \( q = -4\pi\varepsilon_0 \alpha a^2 \) and \( E_c = \frac{\alpha}{2\varepsilon_0} \).

Step by step solution

01

Understand Gauss's Law

Gauss's law for electric fields states that the net electric flux through a closed surface is equal to \( \frac{1}{\varepsilon_0} \) times the charge enclosed. Mathematically, \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \). Here, \(\varepsilon_0\) is the permittivity of free space, \(\mathbf{E}\) is the electric field, and \(d\mathbf{A}\) is a differential area on the closed surface.
02

Determine Charge Enclosed

Consider a spherical Gaussian surface of radius \(r\) such that \(a < r < b\). The charge enclosed \(Q_{\text{enc}}\) is calculated by integrating the volume charge density \(\rho(r) = \frac{\alpha}{r}\) from radius \(a\) to radius \(r\). This gives: \[ Q_{\text{enc}} = \int_a^r \rho(r') 4 \pi r'^2 \, dr' = \int_a^r \frac{\alpha}{r'} 4 \pi r'^2 \, dr' = 4\pi \alpha \int_a^r r' \, dr' \]The integral evaluates to:\[ Q_{\text{enc}} = 4\pi \alpha \left[ \frac{r'^2}{2} \right]_a^r = 4\pi \alpha \left( \frac{r^2}{2} - \frac{a^2}{2} \right) = 2\pi\alpha (r^2 - a^2) \].
03

Apply Gauss's Law

Now apply Gauss's law to the spherical surface at \(r\), with the electric field \(E\) being radial and constant over the surface:\[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0} \]Substituting \(Q_{\text{enc}}\), we have:\[ E \cdot 4\pi r^2 = \frac{2\pi\alpha (r^2 - a^2)}{\varepsilon_0} \]Solving for \(E\):\[ E = \frac{\alpha (r^2 - a^2)}{2\varepsilon_0 r^2} \].
04

Add Point Charge to Make Field Constant

To make the electric field constant for \(a < r < b\), a point charge \(q\) at the center must offset the \(r^2\) dependency. Notice that for a constant field \(E = E_c\), we must have:\[ E_c \cdot 4\pi r^2 = \frac{Q_{\text{enc}} + q}{\varepsilon_0} \].Setting the field to be constant, we equate \[ E_c = \frac{q}{4\pi\varepsilon_0 r^2} + \frac{\alpha}{2\varepsilon_0} - \frac{\alpha a^2}{2\varepsilon_0 r^2} \].For the left terms to be independent of \(r\), the terms at \(r\) cancel by picking \(q = -\alpha a^2 \cdot 4\pi\varepsilon_0\).
05

Determine Constant Electric Field

Once \(q = -4\pi\varepsilon_0 \alpha a^2\), the field is constant and magnitude given by:\[ E_c = \frac{\alpha}{2\varepsilon_0} \].

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Key Concepts

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

Electric Field
The concept of an electric field is fundamental to understand electromagnetic forces. The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any point in space. For any point within the field, the force felt by a tiny test charge can be calculated, which helps us understand how charges interact over a distance.
  • An electric field \(\mathbf{E}\) is typically expressed in units of Newtons per Coulomb (N/C).
  • The direction of the field corresponds to the direction of the force a positive charge would experience.
  • The magnitude depends on the strength of the source charge and the distance from it.

In the context of the hollow sphere problem, Gauss's Law helps in simplifying the calculation of the electric field. It relates the electric field emanating from charges to the total enclosed charge and involves taking advantage of spherical symmetry. The formula derived shows how the electric field depends on the radius \(r\), an aspect that can be controlled by placing additional charges to adjust the field to a constant value.
Charge Density
Charge density \(\rho\) provides a measure of how much electric charge exists in a given area or volume. It is essential for determining how electric fields behave in the presence of distributed charges. In our scenario, we've been given a volume charge density \(\rho(r) = \frac{\alpha}{r}\) within a spherical shell.
  • For volume charge density, the units are typically Coulombs per cubic meter (C/m^3).
  • It describes how charge is distributed in three-dimensional space.
  • The variation with \(r\) gives it a one-dimensional aspect, simplifying the calculations.

Understanding \(\rho(r)\) is crucial for determining how much charge is enclosed within any segment of space. The integral of \(\rho(r)\) over the volume of interest provides the total charge, a fundamental part of applying Gauss's Law effectively. This is particularly applicable in the hollow sphere exercise, as the integration over a spherical volume with radial symmetry simplifies to easier calculations.
Spherical Symmetry
Spherical symmetry refers to a system where all points at a certain radius from the center are identical, making it easier to solve problems involving forces or fields. This symmetry significantly simplifies the calculations of electric fields using Gauss's Law.
  • In spherical symmetry, physical quantities depend only on the distance from the center, not on the direction.
  • The symmetry allows for simplification by converting complex volume integrals into more manageable forms.
  • It is particularly useful in problems like that of the hollow sphere since it reduces the dependency to a single variable (radius).

For the given problem of a hollow spherical charge distribution, we leverage spherical symmetry to construct a Gaussian surface that aligns perfectly with the symmetry, yielding a constant electric field in the desired region. A Gaussian surface in the form of a sphere centered on the distribution makes integrating over the volume straightforward, as the symmetry implies the electric field is isotropic and only a function of radial distance, collapsing a three-dimensional problem into a one-dimensional task.

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Most popular questions from this chapter

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius \(R .\) Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r > R\) from the center of the shell (outside the shell) and (b) a distance \(r < R\) from the center of the shell (inside the shell).

A solid metal sphere with radius 0.450 \(\mathrm{m}\) carries a net charge of 0.250 \(\mathrm{nC}\) . Find the magnitude of the electric field (a) at a point 0.100 \(\mathrm{m}\) outside the surface of the sphere and (b) at a point inside the sphere, 0.100 \(\mathrm{m}\) below the surface.

A point charge of \(-2.00 \mu C\) 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.

At time \(t=0\) a proton is a distance of 0.360 \(\mathrm{m}\) from a very large insulating sheet of charge and is moving parallel to the sheet with speed \(9.70 \times 10^{2} \mathrm{m} / \mathrm{s}\) . The sheet has uniform surface charge density \(2.34 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} .\) What is the speed of the proton at \(t=5.00 \times 10^{-8} \mathrm{s} ?\)

The gravitational force between two point masses separated by a distance \(r\) is proportional to \(1 / r^{2},\) just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss's law for gravitation. (a) Let \(\vec{g}\) be the acceleration due to gravity caused by a point mass \(m\) at the origin, so that \(\vec{g}=-\left(G m / r^{2}\right) \hat{r}\) . Consider a spherical Gaussian surface with radius \(r\) centered on this point mass, and show that the flux of \(\vec{g}\) through this surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G m$$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show that the flux of \(\vec{g}\) through any closed surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G M_{\mathrm{encl}}$$ where \(M_{\text { encl }}\) is the total mass enclosed within the closed surface.
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