Divergence of a filed and its interpretation from images What I know about divergence is like net flux, but from these two examples,
div(Example(A)) =2, which I can see obviously that when I put a little circle anywhere, the length of output is bigger than input.
However, div(Example(B))=0, but from the image, the length of output seems less than input, why does the divergence of this case not less than zero to express the net flow is negative?
The same as div(Example(C))>0, but from the image, I think the net flow is equal to zero?
I'm very confused about this, thank you very much!

Example(A)

Example(B)

Example(C)
 A: I thought I would try and explain as to what is happening without using detailed Mathematical analysis, so mine is really a hand-waving approach.
The net flux is $$\int\int_S \vec E \cdot \vec n\,\, dA$$
So when the calculation of flux is made both the magnitude of the vector and the direction of the vector relative to the direction of the normal to the area have to be taken into account.  
 
The blue lines indicate that there is a flux into the enclosed space and the green lines indicate that there is a flux out of the enclosed space.  
A1 and A2 seem to show a net flux out of the enclosed area  
B1 Here the magnitude of the vectors pointing out might be smaller than the magnitude of the vectors pointing in but there is more surface through which the vectors point outwards than for vectors pointing inwards.
B2 has equal surface area for the vectors pointing in and out but the inclination of the smaller magnitude vectors pointing out relative to the normal to the area is smaller than those of the larger magnitude vectors pointing in.
So overall one could imagine that the net flux is zero for these two examples.  
C1 has more vectors pointing outwards than vectors pointing inwards.
C2 has the same number of vectors pointing in as pointing out but the inclination of the vectors pointing out relative to the normal to the area is smaller than that for the vectors pointing in.
A: First of all, the divergence is indeed like a volumetric density of flux. If you let $\phi$ be the flux and $d\tau$ an infinitesimal volume you have : $$\frac{d\phi}{d\tau} = \mathrm{div}\overrightarrow{F} $$
Something important that you might have missed is that the value of the divergence depends of the place where you are computing it. It is in fact $\mathrm{div}\overrightarrow{F} \! \left(x,y \right)$ .


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*Example A
For this example we find that for any $(x,y)$ we have $\mathrm{div}\overrightarrow{F} \! \left(x,y \right)=2$. So indeed wherever you put a little circle you will have a net positive outgoing flux.


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*Example B
For this one we find that $\mathrm{div}\overrightarrow{F} \! \left(x,y \right)= 0 $. So the divergence is indeed null everywhere the field is defined. The fact that it looks negative is caused by the fact that the arrows with greater length apply on a lesser surface. However, the field is not defined in zero. It tackles infinity; hence a problem. The divergence has no value for the origin.
If you take a circle that is not infinitely small centered in the origin the net outgoing flux flowing through will be positive. If you make it smaller and smaller it will stay positive and even be bigger and bigger : the is due to the fact that the center is a singularity. The divergence has no meaning here 


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*Example C
Here we have  $\mathrm{div}\overrightarrow{F} \! \left(x,y \right)= \frac{1}{\sqrt{x^2+y^2}}$. So here the net outgoing flux is positive, nonetheless if we 
get away from zero the divergence approaches zero. So for human eyes, given that we look far enough from the origin, it will look like the outgoing flux is equal to zero, like you thought and it the farther away the closer to zero it will get. 

