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Given the flux density \(\mathbf{D}=\frac{16}{r} \cos (2 \theta) \mathbf{a}_{\theta} \mathrm{C} / \mathrm{m}^{2}\), use two different methods to find the total charge within the region \(1

Short Answer

Expert verified
Using the given information, explain how you would calculate the total charge within the region \(1<r<2 \mathrm{~m}, 1<\theta<2 \mathrm{rad}, 1<\phi<2 \mathrm{rad}\) by two different methods: using the volume integral of charge density and using surface integral of flux density.

Step by step solution

01

Find the charge density

To find the charge density \(\rho\), we will use the relation between flux density and charge density: \(\mathbf{D} = \epsilon\mathbf{E} = \epsilon \frac{\rho}{\epsilon} = \rho\). So, in this case, we have \(\rho = \frac{16}{r} \cos (2 \theta)\).
02

Convert to cartesian coordinates

In order to perform the volume integral, it's much easier to work with cartesian coordinates. We have: \(\rho = \frac{16\cos(2\theta)}{r}\) Converting spherical coordinates to cartesian coordinates, we have: \(r^2 = x^2 + y^2 + z^2\) \(\theta = \arccos\left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)\) So, \(\rho(x,y,z) = \frac{16\cos\left(2\arccos\left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)\right)}{\sqrt{x^2 + y^2 + z^2}}\)
03

Perform the volume integral

Now, we can find the total charge by integrating the charge density over the given volume: \(Q_1 = \int_V \rho(x, y, z) dV\) The limits of integration will be determined by the given region: \(1 < r < 2\), \(1 < \theta < 2\), and \(1 < \phi < 2\). Calculate this integral to find the total charge within the region. For method 2:
04

Find the surface charge density

Recall that the surface charge density \(\sigma\) is related to the flux density by \(\mathbf{D} = \sigma\mathbf{a}_n\). In our case, since \(\mathbf{D}\) is in the \(\mathbf{a}_\theta\) direction, the surface charge density will be nonzero on the surfaces defined by \(\theta = 1\) and \(\theta = 2\). Hence, we have: \(\sigma_1 = 16\cos(2)\) for \(\theta=1\) \(\sigma_2 = 16\cos(4)\) for \(\theta=2\)
05

Calculate the total charge using surface integrals

Now we can find the total charge using surface integrals over the two surfaces: \(Q_2 = \int_{S_1} \sigma_1 dS + \int_{S_2} \sigma_2 dS\) The limits of integration will be determined by the surfaces corresponding to \(\theta = 1\) and \(\theta = 2\). Calculate these surface integrals to find the total charge within the region. Finally, compare the results from both methods to confirm that they give the same value for the total charge within the specified region.

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Key Concepts

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

Charge Density
Charge density, often denoted as \( \rho \), is a measure of the amount of electric charge per unit volume in a given region of space. In the context of electromagnetic theory, charge density plays a critical role as it influences the electric field around it. The relation between flux density (also known as electric displacement field and denoted as \(\textbf{D}\)) and charge density is given by Gauss's law, which states that \(\textbf{D}\) is equal to \( \rho \) times the permittivity of the medium (\(\epsilon\)).

In our exercise, the provided flux density is in spherical coordinates, and to find the charge density, one can simply compute \(\rho = \frac{16}{r} \cos(2 \theta)\). Understanding charge density is essential because it allows us to calculate the total charge in a region by integrating \(\rho\) over that region's volume. This fundamental concept is crucial when analyzing electromagnetic fields in various geometries.
Volume Integral
A volume integral extends the concept of an integral, which is traditionally used for finding areas under curves, to three dimensions for finding a quantity within a volume. In the realm of physics and engineering, particularly in electromagnetic theory, volume integrals allow us to calculate properties like total charge or mass within a three-dimensional space.

For our problem, the volume integral is used to compute the total charge within a specific region by integrating the charge density \(\rho\) over the entire volume \(V\). The mathematical expression for the total charge \(Q_1\) is given by \(Q_1 = \int_V \rho(x, y, z) \, dV\), where the limits of integration are derived from the conditions defining the region's boundaries. It's crucial for students to be comfortable with setting up and evaluating volume integrals, as it directly applies to many problems in electromagnetic theory.
Surface Charge Density
Surface charge density, denoted by \(\sigma\), is another fundamental concept in electromagnetism, which specifies the amount of electric charge per unit area on a surface. It becomes particularly relevant when dealing with scenarios where charges are distributed over a surface rather than throughout a volume.

In our exercise, the flux density given inherently defines a surface charge density on specific surfaces within the spherical coordinate system. These surfaces are characterized by \(\theta = 1\) rad and \(\theta = 2\) rad. The surface charge densities, \(\sigma_1\) and \(\sigma_2\), on these surfaces are calculated from the flux density components parallel to the surface normal. To find the total charge using surface integrals, one must integrate \(\sigma\) over the surface area \(S\). Understanding and calculating surface charge density allows us to analyze the distribution of charge in more complex geometric configurations and is invaluable for solving a wide array of problems in electromagnetic theory.
Spherical to Cartesian Coordinates Conversion
Converting spherical coordinates to Cartesian coordinates is a fundamental mathematical skill needed to solve spatial problems in physics and engineering. The two coordinate systems offer different benefits, with spherical coordinates often simplifying problems with radial symmetry, while Cartesian coordinates are preferable for volume integrals due to their straightforward differential element.

In solving our exercise, the conversion from spherical coordinates \( (r, \theta, \phi) \) to Cartesian coordinates \( (x, y, z) \) is imperative to set up the volume integral properly. The spherical coordinate \(r\) equates to the square root of the sum of the squares of the Cartesian coordinates, and the angle \(\theta\) is the arccosine of the \(z\)-coordinate over the radial distance. This conversion is applied to express the charge density \(\rho\) in terms of \(x, y, z\), as seen in our exercise. Mastering this transformation is crucial for students to traverse between coordinate systems fluidly when approaching multidimensional problems.
Electromagnetic Theory Concepts
Understanding core concepts in electromagnetic theory is integral to solving exercises like the one provided. Flux density \(\mathbf{D}\), charge density \(\rho\), volume integrals, and surface charge density \(\sigma\) are just a few of the vital elements that students need to be familiar with. Beyond these, one must know about electric fields, Gauss's law, the relationships between electric fields and potential, and how different coordinate systems interact with these fields.

The task of finding the total charge within a given region shows not only the theoretical applications but also practically applies these concepts. A comprehensive grasp of how these elements interrelate allows students to tackle a variety of problems within electromagnetic theory, from basic ones involving uniform fields to more complex scenarios involving dynamic electromagnetic environments. A strong foundation in these concepts is essential for any student or professional working in fields related to electromagnetism.

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Most popular questions from this chapter

A radial electric field distribution in free space is given in spherical coordinates as: $$ \begin{array}{l} \mathbf{E}_{1}=\frac{r \rho_{0}}{3 \epsilon_{0}} \mathbf{a}_{r} \quad(r \leq a) \\ \mathbf{E}_{2}=\frac{\left(2 a^{3}-r^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(a \leq r \leq b) \\ \mathbf{E}_{3}=\frac{\left(2 a^{3}-b^{3}\right) \rho_{0}}{3 \epsilon_{0} r^{2}} \mathbf{a}_{r} \quad(r \geq b) \end{array} $$ where \(\rho_{0}, a\), and \(b\) are constants. \((a)\) Determine the volume charge density in the entire region \((0 \leq r \leq \infty)\) by the appropriate use of \(\nabla \cdot \mathbf{D}=\rho_{v} \cdot(b) \mathrm{In}\) terms of given parameters, find the total charge, \(Q\), within a sphere of radius \(r\) where \(r>b\).

Calculate \(\nabla \cdot \mathbf{D}\) at the point specified if \((a) \mathbf{D}=\left(1 / z^{2}\right)\left[10 x y z \mathbf{a}_{x}+\right.\) \(\left.5 x^{2} z \mathbf{a}_{y}+\left(2 z^{3}-5 x^{2} y\right) \mathbf{a}_{z}\right]\) at \(P(-2,3,5) ;(b) \mathbf{D}=5 z^{2} \mathbf{a}_{\rho}+10 \rho z \mathbf{a}_{z}\) at \(P\left(3,-45^{\circ}, 5\right) ;(c) \mathbf{D}=2 r \sin \theta \sin \phi \mathbf{a}_{r}+r \cos \theta \sin \phi \mathbf{a}_{\theta}+r \cos \phi \mathbf{a}_{\phi}\) at \(P\left(3,45^{\circ},-45^{\circ}\right) .\)

Spherical surfaces at \(r=2,4\), and \(6 \mathrm{~m}\) carry uniform surface charge densities of \(20 \mathrm{nC} / \mathrm{m}^{2},-4 \mathrm{n} \mathrm{C} / \mathrm{m}^{2}\), and \(\rho_{\mathrm{so}}\), respectively. \((a)\) Find \(\mathbf{D}\) at \(r=1\), 3 , and \(5 \mathrm{~m}\). (b) Determine \(\rho_{S 0}\) such that \(\mathbf{D}=0\) at \(r=7 \mathrm{~m}\).

The sun radiates a total power of about \(3.86 \times 10^{26}\) watts \((\mathrm{W})\). If we imagine the sun's surface to be marked off in latitude and longitude and assume uniform radiation, \((a)\) what power is radiated by the region lying between latitude \(50^{\circ} \mathrm{N}\) and \(60^{\circ} \mathrm{N}\) and longitude \(12^{\circ} \mathrm{W}\) and \(27^{\circ} \mathrm{W} ?(b)\) What is the power density on a spherical surface \(93,000,000\) miles from the sun in \(\mathrm{W} / \mathrm{m}^{2} ?\)

A certain light-emitting diode (LED) is centered at the origin with its surface in the \(x y\) plane. At far distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation pattern that follows a raised cosine law: that is, the optical power (flux) density in watts \(/ \mathrm{m}^{2}\) is given in spherical coordinates by $$ \mathbf{P}_{d}=P_{0} \frac{\cos ^{2} \theta}{2 \pi r^{2}} \mathbf{a}_{r} \quad \text { watts } / \mathrm{m}^{2} $$ where \(\theta\) is the angle measured with respect to the direction that is normal to the LED surface (in this case, the \(z\) axis), and \(r\) is the radial distance from the origin at which the power is detected. \((a)\) In terms of \(P_{0}\), find the total power in watts emitted in the upper half-space by the LED; (b) Find the cone angle, \(\theta_{1}\), within which half the total power is radiated, that is, within the range \(0<\theta<\theta_{1} ;\) ( \(c\) ) An optical detector, having a \(1-\mathrm{mm}^{2}\) cross-sectional area, is positioned at \(r=1 \mathrm{~m}\) and at \(\theta=45^{\circ}\), such that it faces the \(\mathrm{LED}\). If one milliwatt is measured by the detector, what (to a very good estimate) is the value of \(P_{0}\) ?

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