How is it physically possible that the electric field of some charge distributions does not attenuate with the distance? Let's consider for instance an infinite plane sheet of charge: you know that its E-field is vertical and its Absolute value is $\sigma / 2 \epsilon _0$, which is not dependent on the observer position.
How is this physically possible? An observer may put himself at an infinite distance from all charges and he will receive the same E-field. It seems strange.
 A: Instead of thinking about the sheet of charge as having infinite dimensions, think of the dimensions being much greater than the distance from the charge where the field is measured. The electric field between the parallel plates of a capacitor is considered uniform (except near the edges) yet the dimensions of the plates are obviously not infinite. They’re just so much greater than the plate separation you can think of them as “infinite” relatively speaking.
The  field from a individual charge varies as the inverse square of the distance from the charge, if you integrate the contribution of each charge from an “infinite” plane of charge the resulting field is constant. A proof can be found here: http://mlg.eng.cam.ac.uk/mchutchon/chargedPlanes.pdf
Hope this helps.
A: The answer is simple.  It is not physically possible.   No charge distribution can extend without limit.  Every charge system is bounded.   If you are farther away from a flat charge distribution than it's size, the electric field will attenuate with distance.
A: First, by symmetry, you would expect the field to be perpendicular to the plane. All the sideways components cancel. 
You can calculate the field at a point by adding up the field from all the charges in the plane. We start with a point a distance $z$ from the sheet. We can get the total field by integrating over rings as shown. We will focus on just one ring at angle $\theta$ and angular thickness $\Delta \theta$. We calculate that the field from this ring is $\Delta E$.
Now we double the distance to $2z$ and look at another ring using the same angles. This ring is twice as far away, but has twice the circumference and twice the thickness. So the field from it is the same $\Delta E$ as before.
If you integrate over all such rings, you can see that the total doesn't change. 

This shows the result is right, but still leaves the uncomfortable feeling that as you get closer to a collection of charges, the field from each one gets bigger. So the total ought to get bigger. Yet that that doesn't happen. 
Consider our original ring from a distance $z$. Now move in to $z/2$ and look at the same ring. We are closer to it, and the field from each element in it is stronger. But the angle has changed. The fields from elements on opposite sides come closer to canceling each other. The total field from that ring is actually smaller at a closer distance. If we were actually in the plane, the field from that ring would be $0$.
A: In general, the field felt at a particular distance from a particular charge distribution is the product of two factors:


*

*How much charge you see in a particular direction (i.e. a small angular area), and

*How far away that charge is.


The second factor is relatively simple - the electric field has a $1/r^2$ dependence, so as you move further away from a charge distribution, it will always shrink by $1/r^2$. The first factor is more interesting.
Let's examine how each of these quantities changes when you move further away from a few example charge distributions:
Point charge:
For a point charge, the amount of charge that you see in any particular direction is constant. The point charge is either in the angular cone that you choose, or it isn't, no matter how far away you are. As we already said, moving further away from the charge distribution gives you an automatic $1/r^2$ decrease in the field magnitude. So, multiplying these two factors together gives you $(\text{constant})\times\frac{1}{r^2}\propto 1/r^2$ dependence on distance.
Line charge:
For a line charge, the total charge on a given segment of the line is proportional to the length of the line segment. As you move further away from a line charge, the length of the line segment that is enclosed in a given angular cone increases linearly with distance (you can think of it as the arc length subtended by that angle, given by $s=r\theta$). Since moving further away causes the length of the line enclosed by a given angle to increase, then the total charge within a given angle increases linearly with distance. As before, the second factor gives you a $1/r^2$ dependence, so, in total, we have $r\times1/r^2\propto 1/r$ dependence for this distribution.
Sheet of charge:
For a sheet of charge, the total charge on a given patch of the sheet is proportional to the area of the patch. As you move further away from the sheet, the area enclosed by each angular cone grows quadratically with distance (we already saw that the width of the region enclosed by the cone grows linearly with distance, and area is proportional to squared width). This means that the total charge enclosed by a given angular cone also increases quadratically with distance. So we have $r^2\times 1/r^2\propto\text{constant}$ dependence on distance.

So, ultimately, the constant field as a function of distance is due to two competing factors exactly cancelling each other out for a very particular configuration of charge. The further away you move, the more charge you see in any given direction, but the further away that charge is. 
A: 
An observer may put himself at an infinite distance from all charges
  and he will receive the same E-field.

I'm compelled to address this misconception just in case it is at the root of your question.
When we solve this problem for the electrostatic field, the result is independent of the distance $r$ from the plane. Your intuition seems to inform you that this isn't possible since an observer can get infinitely far away from all charges. But an observer can't get infinitely far away, an observer can get arbitrarily far away but $r$ must have a value - infinity isn't a number that the value of $r$ can take.
Through a limit process, one can talk meaningfully about the electric field strength as $r$ goes to infinity but I think it's a conceptual error to think in terms of being an infinite distance away from all charges.
My gut tells me that you're imagining that one can get far enough away from the plane that the 'size' of the plane shrinks to zero but that's not the case by stipulation.
To say that the plane of charge is infinite is to say that the plane has no edge. Thus, no matter how large $r$ (the distance from the plane) becomes, there is no edge to be seen.
A: I believe that it is your intuition that is failing here, and since mathematical arguments have already been provided, I'll offer a simpler and more intuitive approach.
Take a more or less similar example which is more familiar to you, the sky. The sky, from where we are, looks like an infinite plane, this is not entirely correct but we can assume that from our point of view and for the sake of the argument. You may notice that you don't see a bigger or smaller sky as you are higher or lower over the ground (lets say as you move up and down inside a building.) This is because the plane we are looking at has actually infinite extension. If it didn't, you may see how its "apparent dimension" changes, as you may experience with any object of your everyday life. Of course, this is just a simple comparison to help intuition which gets often confused when the infinite is involved, not really an argument for the constant field, in which case we have no better approach that the mathematical proof.
