Electromagnetics problem....

[Firefox123]

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

[arildno]

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

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

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

Then we have, by Gauss'theorem:
\int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}=\int_{\ma thcal{V}}\nabla\cdot\vec{A}dV

or, rearranging a bit:
\int_{\mathcal{H}}\vec{A}\cdot{d}\vec{S}=\int_{\ma thcal{V}}\nabla\cdot\vec{A}dV-\int_{\mathcal{D}}\vec{A}\cdot{d}\vec{S}

[Firefox123]

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

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

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

Then we have, by Gauss'theorem:
\int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}=\int_{\ma thcal{V}}\nabla\cdot\vec{A}dV

or, rearranging a bit:
\int_{\mathcal{H}}\vec{A}\cdot{d}\vec{S}=\int_{\ma thcal{V}}\nabla\cdot\vec{A}dV-\int_{\mathcal{D}}\vec{A}\cdot{d}\vec{S}


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 \int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S} directly so that one can get practise writing out the differential surface element.

Since they gave me the vector in cartesian coordinates \vec{A}=\vec{a_z}zI am assuming they also want me to practise converting this to spherical coordinates.

I believe after that I have to find \vec{\nabla} \cdot \vec{A}, 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 \int_{\mathcal{S}}\vec{A}\cdot{d}\vec{S}.....so either my limits are incorrect or the two expressions I am taking the dot product of are incorrect.


Russ

[arildno]

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

Did this clarify anything?

[arildno]

Oh dear, the area element should of course be:
dS=R^{2}\sin\phi{d}\theta{d}\phi

[Chen]

I'm not sure, but aren't you simply asked to find the flux of whatever \vec{A} represents through the hemispherical surface? And in that case, isn't the flux of \vec{A} through the base of the hemisphere equal to the flux through the upper surface? This would lead to:

\phi _A = \vec{A}\cdot{\vec{B}} = A\pi R^2

Where \vec{B} represents the base of the hemisphere, and R is its radius.

[arildno]

I'm not sure, but aren't you simply asked to find the flux of whatever \vec{A} represents through the hemispherical surface? And in that case, isn't the flux of \vec{A} through the base of the hemisphere equal to the flux through the upper surface? This would lead to:

\phi _A = \vec{A}\cdot{\vec{B}} = A\pi R^2

Where \vec{B} represents the base of the hemisphere, and R is its radius.

That would be true if the \nabla\cdot\vec{A}=0 in the interior, by the divergence theorem.

[Firefox123]

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

Did this clarify anything?

Since you corrected the area element.....

Oh dear, the area element should of course be:
dS=R^{2}\sin\phi{d}\theta{d}\phi

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
\vec{A}\cdot{d}\vec{S} ?

***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

[arildno]

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

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

Integrating:
\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\vec{A}\cdo t{d}\vec{S}=2\pi{R}^{3}\int_{0}^{\frac{\pi}{2}}\co s^{2}\phi\sin\phi{d\phi}
=2\pi{R}^{3}(-\frac{1}{3}\cos^{3}\phi)\mid_{0}^{\frac{\pi}{2}}=\ frac{2\pi{R}^{3}}{3}

[Firefox123]

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

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

Integrating:
\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\vec{A}\cdo t{d}\vec{S}=2\pi{R}^{3}\int_{0}^{\frac{\pi}{2}}\co s^{2}\phi\sin\phi{d\phi}
=2\pi{R}^{3}(-\frac{1}{3}\cos^{3}\phi)\mid_{0}^{\frac{\pi}{2}}=\ frac{2\pi{R}^{3}}{3}

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 \vec{a_r} direction and is given by...
dS={R}^{2}\sin\phi{d}\theta{d}\phi\vec{a_r}

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

\vec{A}=R\cos(\phi)\vec{a}_{z}


since we know that z=R\cos(\phi)


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



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

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

[arildno]

No problem :smile:

[Firefox123]

One more question....related to Latex....

And we can rewrite the vector \vec{A}=\vec{a_z}z
as:\vec{A}=\{R}cos\theta\vec{a_z}

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

and also this part....

since we know that
\{Z}={R}\cos\theta

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

And finally...

So know we have :
\vec{A}\cdot{d}\vec{S}=\R\cos\theta\vec{a_z}\cdot\ {R}^{2}\sin\phi{d}\theta{d}\phi\vec{a_r}

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

[arildno]

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

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

Click on the LATEX graphics to reveal the code

[Firefox123]

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

True...I noticed that also. I think sometimes there is a difference between the way that engineering books and physics books define the angles.

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

Click on the LATEX graphics to reveal the code

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




Russ

[arildno]

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 \phi,\theta

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 \theta as the symbol for the angle.
Why should we suddenly redefine the planar angle into \phi when we take the step into 3-D??
To me, this move seems completely unmotivated and fundementally unpedagogical. :grumpy:

[Firefox123]

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 \phi,\theta

Sounds like you have strong feelings about this issue....
:uhh: please dont hurt me.....

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 \theta as the symbol for the angle.
Why should we suddenly redefine the planar angle into \phi when we take the step into 3-D??
To me, this move seems completely unmotivated and fundementally unpedagogical. :grumpy:

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

[arildno]

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..)

[allanpatrick]

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.