Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) A flux density field is given as \(\mathbf{F}_{1}=5 \mathbf{a}_{z} .\) Evaluate the outward flux of \(\mathbf{F}_{1}\) through the hemispherical surface, \(r=a, 0<\theta<\pi / 2,0<\phi<2 \pi\) (b) What simple observation would have saved a lot of work in part \(a ?\) (c) Now suppose the field is given by \(\mathbf{F}_{2}=5 z \mathbf{a}_{z} .\) Using the appropriate surface integrals, evaluate the net outward flux of \(\mathbf{F}_{2}\) through the closed surface consisting of the hemisphere of part \(a\) and its circular base in the \(x y\) plane. ( \(d\) ) Repeat part \(c\) by using the divergence theorem and an appropriate volume integral.

Short Answer

Expert verified
Question: Evaluate the outward flux of the given vector field \(\mathbf{F}_1\) through the hemispherical surface of radius a. Answer: The outward flux of \(\mathbf{F}_1\) through the given hemisphere is 0.

Step by step solution

01

(a) Evaluate outward flux of \(\mathbf{F}_1\) through the hemispherical surface

To evaluate the outward flux of \(\mathbf{F}_1\) through the given hemisphere, we need to compute the surface integral of \(\mathbf{F}_1 \cdot \mathbf{n}\), where \(\mathbf{n}\) is the outward unit normal vector. The surface integral can be written as \(\oint_S \mathbf{F}_{1} \cdot \mathbf{n} dS\) In spherical coordinates, the surface element \(dS = a^2 \sin\theta d\theta d\phi\) and the outward unit normal is \(\mathbf{n} = \mathbf{a}_r\). Thus, we have \(\oint_S \mathbf{F}_{1} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{\pi/2} (5 \mathbf{a}_z \cdot \mathbf{a}_r) a^2 \sin\theta d\theta d\phi\) Since, \(\mathbf{a}_z \cdot \mathbf{a}_r = 0\) (orthogonal unit vectors), the outward flux is 0.
02

(b) Simple observation to save work in part (a)

The simple observation is that since \(\mathbf{F}_1\) is parallel to the z-axis and the hemisphere lies in the xy-plane, the vector field and the outward normal vectors are orthogonal. Hence, their dot product will be zero, resulting in zero outward flux.
03

(c) Evaluate net outward flux of \(\mathbf{F}_2\) through the closed surface

For the closed surface, we have two regions: the hemisphere surface (as given in part (a)) and the circular base lying in the xy-plane. Let's compute the surface integrals for these two regions. 1. Hemispherical Surface: \(\oint_S \mathbf{F}_{2} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{\pi/2} (5z \mathbf{a}_z \cdot \mathbf{a}_r) a^2 \sin\theta d\theta d\phi\) Since, \(\mathbf{a}_z \cdot \mathbf{a}_r = 0\), the outward flux through the hemisphere is 0. 2. Circular base: For the circular base, the outward normal is \(\mathbf{n} = -\mathbf{a}_z\), and the surface element is \(dS = r dr d\phi\). \(\oint_{S_b} \mathbf{F}_{2} \cdot \mathbf{n} dS = \int_0^{2\pi} \int_0^{a} (-5z \mathbf{a}_z \cdot \mathbf{a}_z)r dr d\phi = -5 \int_0^{2\pi} \int_0^{a} r^3 dr d\phi\) Now evaluate the integral to get the flux through the circular base: \(-5 \int_0^{2\pi} \int_0^{a} r^3 dr d\phi = -5 \left[\frac{a^4}{4}\right] \left[2\pi\right]\) Thus, the net outward flux of \(\mathbf{F}_2\) through the closed surface is the sum of flux through the hemispherical surface and circular base, which is \(0 - 5\frac{a^4\pi}{2}\) = \(-\frac{5\pi a^4}{2}\).
04

(d) Use divergence theorem for net outward flux of \(\mathbf{F}_2\)

According to the divergence theorem, \(\oint_S \mathbf{F} \cdot \mathbf{n} dS = \iiint_V (\nabla \cdot \mathbf{F}) dV\) The divergence of \(\mathbf{F}_2\) is \(\nabla \cdot \mathbf{F}_2 = \frac{\partial(5z)}{\partial z} = 5\) Now, evaluate the volume integral of the divergence: \(\iiint_V (\nabla \cdot \mathbf{F}_2) dV = 5 \iiint_V dV\) As the volume is the half of a sphere, \(V = \frac{2}{3}\pi a^3\). Thus, \(5 \iiint_V dV = 5\frac{2\pi a^3}{3}\) Since the net outward flux obtained in part (c) and the volume integral of the divergence is not equal, we cannot use the divergence theorem for the net outward flux of \(\mathbf{F}_2\) through the closed surface consisting of the hemisphere and the circular base.

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 Vaia!

Key Concepts

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

Surface Integral
Understanding surface integrals is like peeling an orange and trying to measure how much peel there is. Just like with the orange peel, in mathematics, we often need to measure things that are spread over a curvy surface, not just a flat one.

In our exercise, when we're looking at the flux of a field \textbf{F} across a surface S, we're essentially trying to add up how much of the field is poking through the surface. To make this more precise, mathematicians created the concept of a surface integral. Think about having a net and measuring how much water flows through it; that's the idea here, but with fields like electromagnetic ones rather than water.

Mathematically, the surface integral of a vector field \textbf{F} across a surface S is written as \[\oint_S \textbf{F} \cdot \textbf{n} dS\], where \[dS\] is a tiny bit of area on the surface and \[\textbf{n}\] is the vector that stands upright on every little piece of the surface, pointing outward. In the case of a constant field like \[\mathbf{F}_{1}\], if it is always perpendicular to the surface, like a gust of wind hitting a wall, then the flow through the surface is simply zero, because the wall doesn't let any wind pass through!
Divergence Theorem
Imagine you have a box, and there is a bunch of confetti shooting out of it — the divergence theorem is like a tool to count all the confetti both inside and as it leaves the box. In more technical terms, the divergence theorem connects the flow of a field out of a closed surface to what's happening inside the volume bounded by the surface.

The divergence theorem says that the total 'stuff' coming out of a closed surface is the same as the total 'creation' or 'destruction' of 'stuff' inside the surface. Here, by 'stuff', we mean whatever the field represents, like the amount of water or energy. Formally, it is written as \[\oint_S \textbf{F} \cdot \textbf{n} dS = \iiint_V (abla \cdot \textbf{F}) dV\], where \[V\] is the volume inside the surface S, \[\abla \cdot \textbf{F}\] is called the divergence of \textbf{F}, and it measures the 'creation' or 'destruction' rate of the 'stuff' inside the volume.

In our exercise problem, we saw that the divergence theorem didn't add up because we dealt with only part of a closed surface — the hemisphere without its base. To apply the divergence theorem correctly, we need a fully enclosed volume. That's why when we closed off the surface with a base in part (c) of the exercise, we could use the theorem to calculate flux by just looking at what happens inside the sphere.
Spherical Coordinates
Navigating using spherical coordinates is like being a pilot in a three-dimensional space. Instead of telling your co-pilot to fly 100 miles east and then 200 miles north — which works well on a flat map — in three-dimensional space, you'd give directions based on how far you should fly, at which angle upward, and what compass direction to take.

Spherical coordinates are a way of describing points in three-dimensional space using three values: the distance from the origin (like the radius of a sphere, called r), the angle down from the z-axis (think of latitude, called \[\theta\]), and the angle around the z-axis (like longitude, called \[\phi\]). They are super useful for problems involving spheres or circular symmetry, just like in our exercise, where we looked at a hemispherical surface.

The transition from x, y, z coordinates to spherical ones (\textbf{r}, \[\theta\], \[\phi\]) is given by the equations \[x = r \sin(\theta) \cos(\phi)\], \[y = r \sin(\theta) \sin(\phi)\], and \[z = r \cos(\theta)\]. Our problem takes full advantage of this system, calculating the flux through a hemisphere, which is much easier to describe with spherical coordinates. Without this coordinate system, dealing with curved surfaces in three dimensions would be like trying to follow a curved path by only taking right angles — very tricky and not very efficient.

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

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 \(\mathrm{W} / \mathrm{m}^{2} ?\)

State whether the divergence of the following vector fields is positive, negative, or zero: ( \(a\) ) the thermal energy flow in \(\mathrm{J} /\left(\mathrm{m}^{2}-\mathrm{s}\right)\) at any point in a freezing ice cube; \((b)\) the current density in \(\mathrm{A} / \mathrm{m}^{2}\) in a bus bar carrying direct current; \((c)\) the mass flow rate in \(\mathrm{kg} /\left(\mathrm{m}^{2}-\mathrm{s}\right)\) below the surface of water in a basin, in which the water is circulating clockwise as viewed from above.

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 \(P_{0}\) ?

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 \(r\).

Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin, \(Q_{1}\), and with hemispheres of radius a. A second charge \(Q_{2}\) (this time a point charge) is located at distance \(R\) from \(Q_{1}\), where \(R>>a .(a)\) What is the force on the point charge before the hemispheres are assembled around \(Q_{1} ?\) (b) What is the force on the point charge after the hemispheres are assembled but before they are discharged? ( \(c\) ) What is the force on the point charge after the hemispheres are assembled and after they are discharged? ( \(d\) ) Qualitatively, describe what happens as \(Q_{2}\) is moved toward the sphere assembly to the extent that the condition \(R>>a\) is no longer valid.

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free