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A cube is defined by \(1

Short Answer

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In summary: a) The total flux leaving the closed surface of the cube is 2.4. b) The divergence of D at the center of the cube is (4.84, 8.476). c) The total charge enclosed within the cube is approximately 0.0216 C.

Step by step solution

01

(a) Apply Gauss's law to find the total flux leaving the closed surface of the cube

To find the total flux leaving the closed surface of the cube, we need to integrate the field D over the six faces of the cube. Let's first consider the four faces perpendicular to the x and y axes: \(i.\) Face with x = 1: \(\int_{1}^{1.2} \int_{1}^{1.2} (2(1)^{2} y) dy dz = 4 \int_{1}^{1.2} \int_{1}^{1.2} y dy dz\) \(ii.\) Face with x = 1.2: \(\int_{1}^{1.2} \int_{1}^{1.2} (2(1.2)^{2} y) dy dz = 6.912 \int_{1}^{1.2} \int_{1}^{1.2} y dy dz\) \(iii.\) Face with y = 1: \(\int_{1}^{1.2} \int_{1}^{1.2} (3x^{2}(1)^{2}) dx dz = 3 \int_{1}^{1.2} \int_{1}^{1.2} x^2 dx dz\) \(iv.\) Face with y = 1.2: \(\int_{1}^{1.2} \int_{1}^{1.2} (3x^2(1.2)^2) dx dz = 4.32 \int_{1}^{1.2} \int_{1}^{1.2} x^2 dx dz\) The electric displacement field has no component in the z direction, so there will be no flux through the faces with z = 1 and z = 1.2. Now, the total flux through the surface is the sum of the fluxes through each face: \(\Phi = \int_{1}^{1.2} \int_{1}^{1.2} (4 + 6.912 + 3x^2 + 4.32x^2) dy dz\) Evaluating this integral, we get: \(\Phi = 2.4\)
02

(b) Evaluate the divergence of D at the center of the cube

To evaluate the divergence of D at the center of the cube, we first need to calculate the divergence of D: \(\nabla \cdot \mathbf{D} = \left(\frac{\partial}{\partial x} (2x^2y), \frac{\partial}{\partial y} (3x^2y^2)\right) = (4xy, 6x^2y)\) The coordinates of the center of the cube are \((1.1, 1.1, 1.1)\). Substituting these values into the divergence of D: \(\nabla \cdot \mathbf{D}\Big|_{(1.1, 1.1, 1.1)} = (4(1.1)(1.1), 6(1.1)^2(1.1)) = (4.84, 8.476)\)
03

(c) Estimate the total charge enclosed within the cube using Equation (8)

To estimate the total charge enclosed within the cube, we need to integrate the divergence of D over the volume of the cube: \(Q = \int_{1}^{1.2} \int_{1}^{1.2} \int_{1}^{1.2} \nabla \cdot \mathbf{D} dV\) \(Q = \int_{1}^{1.2} \int_{1}^{1.2} \int_{1}^{1.2} (4xy + 6x^2y) dy dx dz\) Evaluating this integral, we get: \(Q = 0.0216 C\) So the total charge enclosed within the cube is approximately 0.0216 C.

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Key Concepts

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

Electrical Flux
Electrical flux, a cornerstone of electromagnetics, refers to the amount of electric field that passes through a given area. It is symbolically represented as \text\textGreek\textPhi_{E}\text and can be vividly imagined as the number of electric field lines penetrating a surface. In mathematical terms, the electric flux through a surface S is defined by the integral of the electric field \( \textbf{E} \) over this surface, represented as \( \text\textGreek\textPhi_{E} = \textintolimits_S\textbf{E} \textcdot d\textbf{A} \), where \( d\textbf{A} \) specifies an infinitesimal area on the surface with a direction normal to the surface.

In the context of Gauss's Law, this concept is essential as it equates the net flux through a closed surface to the charge enclosed by the surface, divided by the permittivity of the space. For our cube problem, calculating the flux required integrating the given electric displacement field \( \textbf{D} \) over the six faces of the cube, considering the direction and magnitude of the field on each face.
Divergence of Electric Field
Divergence in electromagnetics is a measure of the net 'outflow' of an electric field from an infinitesimal volume at a given point in space. Mathematically, it is denoted as \( abla \textcdot \textbf{E} \) for the electric field \( \textbf{E} \). In essence, it tells us whether a particular region behaves like a 'source' or a 'sink' of the electric field.

The divergence of the field is a key factor in Gauss's Law for electricity, which is fundamentally linked to charge distribution. In our cube exercise, the divergence was calculated at a specific point - the center of the cube. After finding the partial derivatives of the electric displacement field \( \textbf{D} \) concerning each coordinate, these values were substituted to get the divergence at that central point. This step is vital for understanding how the electric field behaves at different points in space and is used in determining the charge density within a particular region.
Enclosed Charge Estimation
The estimation of the enclosed charge in a volume is an important application of electromagnetics that follows from Gauss's Law. It states that the total charge enclosed by a closed surface can be found by integrating the divergence of the electric field throughout the volume enclosed by the surface, or in the case of a given electric displacement field \( \textbf{D} \), by \( Q = \textint abla \textcdot \textbf{D} dV \).

For our cube problem, the process involved setting up a triple integral of the divergence of \( \textbf{D} \) over the volume of the cube to estimate the total charge inside it. The result of such an integral provides us with a numerical value of charge, influencing the electric and magnetic behavior of the system. Significantly, this concept helps in understanding how charges distribute in space and affect the surrounding electric field.
Electromagnetic Theory
Electromagnetic theory encompasses the study of electric and magnetic fields and their interactions with matter. At the heart of this theory are Maxwell's equations, which describe how electric charges and currents produce electric and magnetic fields, and conversely, how these fields interact with charges and currents.

One of Maxwell's equations is Gauss's Law, the focal point of our cube exercise. This law fundamentally relates electric fields, charge distributions, and the concept of an electric flux. Whether you are calculating the flux through a surface, evaluating the divergence of an electric field, or estimating the enclosed charge within a volume, you are essentially applying principles of electromagnetic theory to solve real-world problems. This theory is not only foundational for understanding the cube exercise but is also the backbone of modern electrical engineering, informing everything from telecommunications to power generation systems.

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Most popular questions from this chapter

The cylindrical surface \(\rho=8 \mathrm{~cm}\) contains the surface charge density, \(\rho_{S}=\) \(5 e^{-20|z|} \mathrm{nC} / \mathrm{m}^{2} .(a)\) What is the total amount of charge present? \((b)\) How much electric flux leaves the surface \(\rho=8 \mathrm{~cm}, 1 \mathrm{~cm}

An infinitely long cylindrical dielectric of radius \(b\) contains charge within its volume of density \(\rho_{v}=a \rho^{2}\), where \(a\) is a constant. Find the electric field strength, \(\mathbf{E}\), both inside and outside the cylinder.

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 \(r>b\).

Use Gauss's law in integral form to show that an inverse distance field in spherical coordinates, \(\mathbf{D}=A a_{r} / r\), where \(A\) is a constant, requires every spherical shell of \(1 \mathrm{~m}\) thickness to contain \(4 \pi A\) coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with \(r\).

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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