divergence and electric flux density

[PG1995]

Hi

Could you please help me with these queries? Thank you.

Regards
PG

Reference(s):
1: https://www.electro-tech-online.com...vector-concepts-divergence-curl-gradient.html
2: https://www.electro-tech-online.com/threads/divergence-problems.132669/?highlight=divergence

 

[steveB]

For Q1, the charge density has to be infinite at the origin because it is a point charge located at the origin. Charge density is the limit of Q/V as V goes to zero. Since Q is constant and the V is going to zero, then charge density must go to infinity. This is true of all point charges. Really charge density is a description best used for the distribution of many point charges with charge so small that we can assume a continuum approximation. If we have large point charges, then we can use Coulomb's law and do summations rather than integrals.

By the way, Coulomb's law and Gauss' Law are essentially the same law.

I'm not sure I'm following you on Q2. All I can say here is that for a point charge at the origin, the charge density has to be zero everywhere away from the origin. This is just the definition of a point charge.

 

[MrAl]

Hello again PG,


The main idea here is not to simply compare the lengths of the arrows, but to compare the arrows to a circle (2d simplification) drawn anywhere inside the field. So it is not a matter of just looking at the lengths of the arrows, you also have to consider the size of a small circle with respect to those arrows. Also, the arrows have direction as i am sure you know, but also the circle itself has direction. The direction in this case is a little unusual...the direction is outward from the center of the circle. That makes it a little strange i know, but that's because it is a unit circle with outward normals. The UCON (Unit Circle with Outward Normals) is the measuring device along with the lengths of the arrows, not just the lengths of the arrows themselves. So when we measure the lengths of the arrows going into the UCON and going out of the UCON we see a slight difference in the total vector sum and thus non zero divergence. But as we make the UCON itself get smaller and smaller, the slight difference gets even smaller, and eventually reaches zero. Thus we finally end up with zero divergence at any point except at the origin.
So we use four mechanisms here (2d simplification):
1. The UCON as the measuring tool.
2. Length of the arrows entering and leaving the UCON.
3. The direction of the arrows with respect to the UCON outward normals (changes the vector used in the sum).
4. The limiting process making the UCON area go to zero.

I believe i tried to show this in detail in the following post. To understand this you'll have to read that in detail because this is a little bit of a tricky subject. It's not that difficult, but it is tricky so you'll have to read it carefully. And i dont think you will find a better complete explanation of this on the web, but just in case you do please post it and i'll be happy to read it too. There may be a fractal proof somewhere too.

See post #7 of this thread:
https://www.electro-tech-online.com/threads/vector-concepts-of-divergence-curl-and-gradient.130663/

If Steve would care to chime in a little perhaps he can add some insight too, and double check my work in that post.

[note this post has been written before Steve posted his post as it took me some time to look up that thread, so Steve beat me by some minutes to the first reply]

 

[PG1995]

Thank you, MrAl.

[note this post has been written before Steve posted his post as it took me some time to look up that thread, so Steve beat me by some minutes to the first reply]

But I had referenced that thread in my first post above! You could have saved yourself some minutes and could have beaten Steve by some minutes to the first reply.

Regards
PG

 

[MrAl]

Hi PG,

He he, yes, that's true. But did that post help at all? Do you understand this better now or do i need to update or something?

 

[steveB]

PG,

I think MrAl provided a very good description and used a 2D description to help you. I think this is a good idea.

I see that my answer was not helpful, probably because I answered from the point of view of what charge and charge density are and not from the point of view of what the divergence theorem is telling you. .... My bad!

Once you have a firm grip on MrAl's examples, I can try to offer a way to see the answers in 3d using the spherically symmetric point charge Q at the origin.

For Q1, I hope you see that the divergence expression is indeterminate at the origin, you have 0/0 because the derivative goes to zero and the r^2 in the denominator (outside the derivative) goes to zero. If indeterminate, then you need to evaluate by a limiting process. Well, we know the answer already, so you actually don't need to go through this effort, but you can do it if you like.

For Q2, remember that the divergence theorem applies to all shapes, whether circles in 2D or spheres in 3D, or any other shape in 3D etc. However, you can make a useful shape to apply the divergence theorem over by considering two spheres of different radii, both with centers at the origin. Then take a particular solid angle and extend it from the origin to infinity. This solid angle will intersect the two spheres and you can make a useful enclosed volume with sides created by the cone created by the solid angle and the two spheres.

This enclosed volume is one you can apply the divergence theorem on in a very easy way. The value of D normal to and over the spherical sections is obvious and there is no normal component of D on the cone section of the surface, so the surface integral is zero there. Hence, the D vector enters into the volume on the inner sphere surface and the D vector exits the volume on the outer sphere surface. Do the calculations there and you will see that two surfaces have equal and opposite integrals.

You mentioned that the vector get smaller at larger radius, which is true. However, the area is getting bigger at the same rate.

Now, it may not be intuitively obvious that the divergence is zero for any surface that does not enclose the point charge, but this is the whole reason why we like the divergence theorem. It tell us a useful truth that might not have been obvious to us at first.

 

[MrAl]

Hello again Steve and PG,

Steve:
PG has told me that he dosnt have the time right now to go over the extended two dimensional example i provided, so i will make a lame attempt here to simplify the results


Simplifying the 2d circle with outward unit normals used as the measuring tool approach...

The circle with outward unit normals makes this problem a little more difficult because normally we measure things with just one device and with just one mechanism, such as placing a ruler along side an object to get the length. But some measurements require a more sophisticated measuring tool and there is no way around this sometimes, and the measuring mechanism is sometimes more complex too. But just as we can measure DC power by multiplying the voltage times the current, we can measure divergence by multiplying each field vector by a unit normal vector where that unit normal vector is found at a position along the circumference of a small circle where the field vector crosses that circumference, and then summing the products of every field vector and unit normal vector.

We have to use a two dimensional circle with outward unit normals as the measuring tool as that is a simplification of the three dimensional sphere with outward unit normals which is a little harder to visualize and certainly harder to draw. There's no simpler view i dont think, because we have to take the direction of the field into account as it is relative to the measuring circle and this shows up as "F*n" in the calculation for the divergence, where F is the field vector and n is the unit normal vector and that is just the dot product of the two. Simplifying that too though, we find that the main point is that the unit normals change the sum over all vector products because it makes some contributions opposite to the others, and thus reduces the overall sum. So when we see one field vector entering the circle with a certain magnitude we see another vector leaving the circle with the same sign and direction, yet we see the vector products sometimes with different sign of that of the field vectors so that the product upon leaving is now opposite to the product upon entering, and since the magnitude of each product difference using a small circle is very small we see a net sum of very close to zero. And as we let the circle area grow smaller and smaller (in order to reach the limiting sum ie the integral) we also see the magnitudes entering and leaving becoming closer and closer to the same value, so eventually the two products cancel out and we are left with zero. So in effect with a very small circle for every vector we find entering the circle we see an equal vector leaving, and because the ones leaving create a product with the directed circle that is opposite in sign to the products entering, they cancel. So we end up with zero divergence

There's a little more to it than that because the field vectors can enter the circle at many different directions and leave at directions that are not the same, but as the circle gets smaller and smaller this effect goes away too unless the divergence is not zero. This can be approximated numerically using real discrete vectors and it would be noted that the more vectors used and the smaller the circle is allowed to become, the closer the approximated divergence becomes equal to zero. I have a 2d example of this too somewhere i think posted in this forum too. Of course if the divergence is not zero (a different field perhaps) then the approximation becomes closer and closer to the exact value of the divergence.

 

[steveB]

I understand if PG is short on time with a school schedule. I guess I would ask him if he still has any basic questions related to this. Were his questions answered sufficiently or does he have any more questions?

I hope my 3d example was clear, but if not we can make a 2D version of that using circles angles and arcs etc. The resulting partial wedge shape is easier to evaluate over compared to a circle because the vector from a point source is either perpendicular or parallel to the perimeter, at all points along it.

 

[PG1995]

Thank you very much, Steve, MrAl, for always being there to help and guide me.

Right now I just wanted to clear one thing here.

PG has told me that he dosnt have the time right now to go over the extended two dimensional example i provided, so i will make a lame attempt here to simplify the results

MrAl: In that PM I didn't say that I didn't have time to go over the extended 2D example.... That's an accusation! Rather I was pointing out that you yourself advised me to read your reply from the other thread. On the other side, yes, my final exams are just around the corner and there is too much material to cram up, and further I had a very important wedding to attend.

Best wishes
PG

 

[MrAl]

Hi PG,

I am not sure what you mean here. You said you didnt have time because you had other things going, but if you do then that's just fine

Also, i like Steve's idea of using the solid angle (3d view) or pie slice (2d view) to explain the divergence too for certain types of fields where these shapes allow more use of symmetry. I use a sphere (3d) or circle (2d) because they are simple shapes with easily calculated borders and this technique is very general and you can move the sphere or circle around like a little magnifying glass to view the div in any part of the field, so it's like using a magnifying glass and thinking, "OK, what is happening over here, and what is happening over here", as you move the circle around within the field.
As others have noted, a warning here is that the div is often hard to view from a graphing of the field because the arrows are too misleading. The arrows show direction and magnitude, but that's it. To know the div we have to know what is happening at a single point and that is very hard to discern from the field of arrows alone without some other mechanism applied to help come up with the right value for the div.

 

[MrAl]

Hello again Steve,


This is funny but i think we may have interpreted PG's question a little incorrectly even though he didnt complain (yet). Here's what it looks like...

For a point charge in cartesian coordinates they get Div D non zero, but in spherical coordinates they show D being zero. I think that is the right way to look at the question.
I had to look up the conversion to spherical coordinates as it's been a while for me, but all that i found seem to agree to what they show there.

Any ideas here?

 

[steveB]

Hi MrAl,

I'm not sure I follow you. Which question might we have miss-interpreted, - Q1 or Q2?

Shouldn't divergence be something that is coordinate independent? Or, are you saying this is true, but the books says the opposite?

 

[MrAl]

Hi Steve,


Sorry about that. What happened was i re-read it and for some reason it struck me as being different somehow. It appeared that they are stating the the Div is non zero (pv) and then later changing that to say that the Div is zero.
Really the correct interpretation is that the first statement is a general statement about the charge, then the later statement (in spherical coord's) is a specific calculation with a specific field. Funny in a way. Probably that came from trying to do too many things in one day with eyes half closed
Thanks for the reply in any case.

 

[Ratchit]

PG,

Q1:

In my opinion, Pv is zero everywhere, at least this is what the divergence expression tells us.

No it does not. Look again. It says explicitly that the divergence expression is only defined for r≠0 .

I don't understand where it says, Pv = 0 everywhere except at the origin, where it is infinite.

Take the case where the infinitesimal volume of a point is not at the origin, and the divergence expression is defined. The same amount of flux will enter the origin side of the infinitesimal volume as leaves from the side away from the origin. That makes it a net flux increase/decrease of zero, and therefore the divergence will be zero. So that checks out.

Now, look at the point of the origin. The divergence expression degenerates to an undefined 0/0, and it is specifically undefined and unusable at the origin. But, we know that the charge Q is constant and cannot be destroyed. So as the infinitesimal volume goes to zero at the origin, we have the expression Q/Δv , which is infinity. So the author of the text is correct after all.

Q2:

Suppose the charge is located at the origin. As the radial distance from the origin increases, the field strength decreases. .... I would say that divergence is zero only when there was no change in the arrow's lengths as the radial distance increases.

You are confused. Electric field intensity is a vector and divergence is a scalar. It does not matter what the field intensity is. The divergence is the net amount of vector flux that enters/leaves an infinitesimal volume regardless of how strong the flux is. So ignore the arrows.

Ratch

 

[MrAl]

Hello,


Also, the projection of a three dimensional field onto a two dimensional surface does not necessarily have the same Div anyway. This tells us right away that the arrows drawn in two dimensional space (from a three dimensional field) may not mean much because the Div may not be the same.

Case in point, the point charge, is just such a field. It's simple projection onto the 2d xy plane changes the Div from zero to some significant non zero value (negative). This means we cant even *look* at the 2d representation (the 2d graph) and expect to see anything that tells us that the Div is zero in the real life 3d case because there is just not enough information there anymore that allows us to calculate the correct Div, even if we take great pains to do so and use advanced methods of some type. Thus any graph of a 3d field in the 2d plane will always come into question about what exactly it represents in relation to the true 3d field...a graph of amplitudes perhaps, but wont represent the Div properly.

 

[MrAl]

Hello again,


Nothing tells a story like a picture sometimes, so here are drawings of two vector fields. The red dots at the ends of the green lines would be the arrow heads if we drew the vectors with arrow heads. If you zoom in a little you can see this better.

The big question is, can you tell which vector field has a divergence of zero?

Note how they are both very similar. They both have vectors who's lengths decrease as the distance from the origin is increased. The one on the right has vector lengths that decrease a little faster than the one on the left, but that's all we can really tell. One of these fields has a divergence of zero and the other one is non zero.

It's just too hard to tell from the vector drawings which one is which and that shows how hard it is too tell. In fact, it's just impossible and here is why...

In two dimensions, one field has zero divergence and the other has negative divergence. But if one of these fields was known to be a cross section of a particular three dimensional field we can say that the divergence is zero of both fields! That's because in 3d that one field has zero divergence, but the field drawn in 2d has negative divergence. So as drawn, one has zero and one has non zero divergence, but the true fields that they come from both have zero divergence. So we can see how drawings can be very deceiving. If we go by the drawing alone, we can be completely wrong about our guess of what the divergence might be.

In case you havent guessed by now, one field is a 2d slice of the point charge electric field. The other field is a 2d field found by a using little guesswork to get the algebraic components that lead to a field which is very similar to the point charge field in 2d and then solving for one of the variables that forces the divergence to equal zero.