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Electromagnetics problem

  1. Jun 2, 2004 #1
    Electromagnetics problem....

    Hello. I am an electrical engineer who is trying to improve my skills in several areas, one of which is electromagnetics.

    I am using the book by David Cheng "Fundamentals in Engineering Electromagnetics" and doing some problems in that book.

    I am having trouble with a problem in chapter 2 of that book (its problem 2-23 or 2-25 I believe)....anyway here is the problem...

    ***Note on my notation: I use bold to signify a vector, since I cant use an arrow on top of it. I also use the symbol "*" for the vector dot product***

    Given a vector A(z) z and a hemispherical surface centered at the origin with radius 3 with the flat bottom in the xy plane find...

    The surface integral of A(z) * ds.

    I have tried it several ways and cant seem to get the answer....Maybe I am writing out the differential surface element incorrectly or my spherical expression for A(z) is wrong.

    If someone could give me a worked out solution I would greatly appreciate it. I normally would not go on the internet for a solution, but I am an Army Reservist currently serving in Kuwait....so my resources for such things are somewhat limited.

    Thanks in advance for any help.



    Russ
     
  2. jcsd
  3. Jun 2, 2004 #2

    arildno

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    Hi, welcome to PF!
    If you are familiar with the LATEX formatting system, this is available at PF;
    check out the thread:
    https://www.physicsforums.com/showthread.php?t=8997

    As to your problem, have you tried using Gauss' theorem?

    Let [tex]\mathcal{S}[/tex] be the closed surface around volume [tex]\mathcal{V}[/tex] consisting of the hemisphere [tex]\mathcal{H}[/tex] and the disk [tex]\mathcal{D}[/tex] in the xy-plane (i.e., z=0).

    Then we have, by Gauss'theorem:
    [tex]\int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}=\int_{\mathcal{V}}\nabla\cdot\vec{A}dV[/tex]

    or, rearranging a bit:
    [tex]\int_{\mathcal{H}}\vec{A}\cdot{d}\vec{S}=\int_{\mathcal{V}}\nabla\cdot\vec{A}dV-\int_{\mathcal{D}}\vec{A}\cdot{d}\vec{S}[/tex]
     
  4. Jun 2, 2004 #3

    Hey arildno....

    First of all thanks for the welcome :smile:

    I am not familiar with LATEX but I will check it out....thanks for the link.

    As to the problem.....I have not used Gauss' theorem because at this point in the problem I dont think the text is looking for that.

    I believe the point of the problem is to solve [tex]\int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}[/tex] directly so that one can get practise writing out the differential surface element.

    Since they gave me the vector in cartesian coordinates [tex] \vec{A}=\vec{a_z}z[/tex]I am assuming they also want me to practise converting this to spherical coordinates.

    I believe after that I have to find [tex] \vec{\nabla} \cdot \vec{A}[/tex], but that seems trivial since I end up taking the partial derivative of z with respect to z and get 1.

    After that I believe I have to verify the divergence theorem by finding the volume integral of the surface, with r=3, but that doesnt seem too difficult since I know what the differential volume element is in spherical coordinates.

    So the main problem right now is writing out the surface integral for [tex]\int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}[/tex].....so either my limits are incorrect or the two expressions I am taking the dot product of are incorrect.


    Russ
     
    Last edited: Jun 2, 2004
  5. Jun 2, 2004 #4

    arildno

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    Ok, let's look at a typical representation of the hemisphere:
    [tex]x=R\sin\phi\cos\theta,y=R\sin\phi\sin\theta,z=R\cos\phi[/tex]
    [tex]R=3,0\leq\theta\leq{2\pi},0\leq\phi\leq\frac{\pi}{2}[/tex]
    [tex]\vec{n}=\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]
    [tex]dS=R^{2}\sin\phi[/tex]

    Did this clarify anything?
     
  6. Jun 2, 2004 #5

    arildno

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    Oh dear, the area element should of course be:
    [tex]dS=R^{2}\sin\phi{d}\theta{d}\phi[/tex]
     
  7. Jun 2, 2004 #6
    I'm not sure, but aren't you simply asked to find the flux of whatever [tex]\vec{A}[/tex] represents through the hemispherical surface? And in that case, isn't the flux of [tex]\vec{A}[/tex] through the base of the hemisphere equal to the flux through the upper surface? This would lead to:

    [tex]\phi _A = \vec{A}\cdot{\vec{B}} = A\pi R^2[/tex]

    Where [tex]\vec{B}[/tex] represents the base of the hemisphere, and R is its radius.
     
  8. Jun 2, 2004 #7

    arildno

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    That would be true if the [tex]\nabla\cdot\vec{A}=0[/tex] in the interior, by the divergence theorem.
     
  9. Jun 2, 2004 #8
    Since you corrected the area element.....

    We are in agreement. I am getting the same expressions that you listed, yet when I take the surface integral of the dot product my answer comes out incorrect.

    So I guess my first question should be....what are you getting for
    [tex]\vec{A}\cdot{d}\vec{S}[/tex] ?

    ***By the way...the book has the answer as (2*pi*R^3)/3

    ***
    I tried using Latex to represent the fraction but I kept on screwing it up so I think I need a bit more practise :tongue2:



    Russ
     
    Last edited: Jun 2, 2004
  10. Jun 2, 2004 #9

    arildno

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    I'm also getting [tex]\frac{2\pi{R}^{3}}{3}[/tex] (:tongue:)

    We have:
    [tex]\vec{A}\cdot{d}\vec{S}=R\cos\phi\vec{k}\cdot\vec{i}_{r}R^{2}\sin\phi{d\phi}{d\theta}[/tex]
    [tex]=R^{3}\cos^{2}\phi\sin\phi{d}\phi{d}\theta[/tex]

    Integrating:
    [tex]\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\vec{A}\cdot{d}\vec{S}=2\pi{R}^{3}\int_{0}^{\frac{\pi}{2}}\cos^{2}\phi\sin\phi{d\phi}[/tex]
    [tex]=2\pi{R}^{3}(-\frac{1}{3}\cos^{3}\phi)\mid_{0}^{\frac{\pi}{2}}=\frac{2\pi{R}^{3}}{3}[/tex]
     
    Last edited: Jun 2, 2004
  11. Jun 2, 2004 #10
    Thanks man. I see the mistake I made...I looked over my work and I never combined two things I had written down on seperate pages. Let me explain my method of doing the problem (Ill point out the part I screwed up)

    The differential area vector for the hemispherical surface is in the [tex]\vec{a_r}[/tex] direction and is given by...
    [tex]dS={R}^{2}\sin\phi{d}\theta{d}\phi\vec{a_r}[/tex]

    And we can rewrite the vector [tex] \vec{A}=\vec{a_z}z[/tex] as:

    [tex]\vec{A}=R\cos(\phi)\vec{a}_{z}[/tex]


    since we know that [tex]z=R\cos(\phi)[/tex]


    So know we have :[tex]\vec{A}\cdot{d}\vec{S}=R\cos(\phi)\vec{a}_{z}\cdot {R}^{2}\sin\phi{d}\phi{d\theta}\vec{a}_{r}[/tex]



    And since the dot product of the direction vectors [tex]\vec{a_z}\cdot\vec{a_r}=cos\theta[/tex]

    We get the expression for the dot product you have above. I was forgetting to include the dot product for the two direction vectors (even though I wrote the damn thing down on another page) and so I was missing a cosine.

    LOL I feel like an idiot.......but anyway.... thanks for your time. Sorry to bother you guys with something so trivial.

    Sometimes I just make dumb mistakes and I dont catch them. *sigh*

    Thanks again.



    Russ
     
    Last edited: Jun 3, 2004
  12. Jun 2, 2004 #11

    arildno

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    No problem :smile:
     
  13. Jun 2, 2004 #12
    One more question....related to Latex....

    That should read "R cos (theta)".... with no bracket

    and also this part....

    That should read "z=R cos (theta)"....with no bracket

    And finally...

    There should be an "R" before the cosine and no bracket before the R^2..

    How do I edit the code to fix these so my post is correct?


    Thanks.


    Russ
     
    Last edited: Jun 2, 2004
  14. Jun 2, 2004 #13

    arildno

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    First of all, you should use [tex]\phi[/tex] rather than [tex]\theta[/tex] in order to have a consistent notation:

    [tex]\vec{A}=R\cos(\phi)\vec{a}_{z}[/tex]
    [tex]z=R\cos(\phi)[/tex]
    [tex]\vec{A}\cdot{d}\vec{S}=R\cos(\phi)\vec{a}_{z}\cdot{R}^{2}\sin\phi{d}\phi{d\theta}\vec{a}_{r}[/tex]

    Click on the LATEX graphics to reveal the code
     
  15. Jun 3, 2004 #14
    True...I noticed that also. I think sometimes there is a difference between the way that engineering books and physics books define the angles.

    Ah. Thanks. Ill try to edit the post and make corrections.




    Russ
     
  16. Jun 3, 2004 #15

    arildno

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    I know, but it is not primarily a difference between physics and engineering books, but 2 "schools" of thought.

    Personally, I abhor using the alternate convention of [tex]\phi,\theta[/tex]

    The reason for my disgust of this practice, is that it is inconsistent with the dominant convention for representing 2-D polar coordinates.
    There, almost without exception, people use [tex]\theta[/tex] as the symbol for the angle.
    Why should we suddenly redefine the planar angle into [tex]\phi[/tex] when we take the step into 3-D??
    To me, this move seems completely unmotivated and fundementally unpedagogical. :grumpy:
     
  17. Jun 4, 2004 #16
    Sounds like you have strong feelings about this issue....
    :uhh: please dont hurt me.....

    Hmmmm.....a friend of mine who was a math/cs major has similar complaints about the engineers notation of spherical coordinates.

    I could see where having two different notations could be somewhat confusing indeed. Why dont engineers and mathematicians/physists use the same notation? Why the difference?



    Russ
     
  18. Jun 4, 2004 #17

    arildno

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    I think it is an example of the inertia of tradition

    Somewhat similarly, I think it is an unnecessary complication that divers practitioners have not yet agreed upon adhering to a single system of units of measure.
    That unit systems other than SI units are still in use, is in my opinion deplorable.

    It cannot be doubted, that a unit system based on the decimal position system is more efficient.
    Why should we suddenly, when doing physics, change the "natural" units appearing in maths?
    Why should, for example a quantity of 12 base units be a new unit, rather than letting a quantity of 10 base units correspond to a new unit?

    For example, when learning maths and units, why should kids be given the additional burden of making (for them) non-trivial conversions after their main computational work?

    If for example they are to add together the lengths of 7 rods, each with lengths 4.8 inches, the will have to rearrange their answers into groups of twelves, to find the number of feet, and so on
    (Isn't it 12 inches to a foot? I never liked those conversions..)
     
    Last edited: Jun 4, 2004
  19. Jun 27, 2004 #18
    help! does anyone have access to Hayt and Buck's solutions manual to Engineering Electromagnetics 6/E? my classmates and i are really faring badly... for anyone who has access to this text's solutions manual, please e-mail me at bluspels_and_flalanspheres@yahoo.com... thanks in advance.
     
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