Interpretation of results, for example those given by Gauss's law After some calculus using Coulomb's law (or more intuitively) or using Gauss' law, we have the results that the field of an infinite line and plane are given respectively by $$\ \ \ \varepsilon_L = \frac { \lambda } { 2 \pi r \epsilon_0} \mathrm{\ \ and \ \ } \varepsilon_P = \frac{ \sigma } { 2 \epsilon_0} $$
What are the intrinsic geometric properties of the infinite line and the inverse square law that would have allowed us to come to this result by inspection? In other words, for the sake of avoiding circular reasoning, could we explain this result without calculus (and hence Gauss's law)? 
Additionally and probably more importantly, what are some strategies that you use to make sense of your results? How often do you engage in this kind of exercise while learning? Is it worth the time? Often, I'll spend tons of time trying to gain insight for results like these, but very rarely arrive at a result.
 A: In the case of an infinite plate, it is possible to make some kind of argument.
Because it is infinite, there is no "natural" length scale for height in this problem, in the sense that, if we compare the situations at some heights $h$ and $2h$ above the plate, then we can find no difference: if we closed our eyes and moved to another height, then reopened our eyes, nothing would have visibly changed in the physics of the problem. Thus we can anticipate that the field would remain the same at $h$ and $2h$.
This argument is somewhat suspicious as it should apply equally well to the wire, for which the field is not constant.  
Geometrically, we go from point charge with field $\sim 1/r^2$ to line charge with field $\sim 1/r$ to infinite plane with field in $~1/r^0=$ constant, which certainly suggest some link to the co-dimension but I have not read a complete argument on this.
A: I just want to pull @MarkH's comment into a full answer because it is important but also importantly incomplete, it has geometric factors that you cannot work out without calculus but a good physical picture.
The Coulomb law says that the force goes like $1/r^2,$ and if you wanted to "turn the tables" on the particle you could imagine it like this: a particle projects all of the charges it sees onto a local spherical screen sitting at a fixed radius $R$ around it, diminishing those charges with distance, so that it can then experiences the same force, but due to the charges on the screen. The idea of "diminishing" the charges of course means multiplying by $R^2/r^2$ where $r$ is the actual distance from the sphere, but this has a nice geometrical interpretation: smear out the charge over some little ball of volume $V_\epsilon = \frac43\pi \epsilon^3,$ so that it is defined as a very large charge density $\rho_\epsilon = Q/V_\epsilon.$ The area of this ball's projection on the sphere is $\pi \epsilon^2 ~ \frac{R^2}{r^2},$ and because we usually don't care about the exact value of $R$ we informally set $R=1$ and call this area the "solid angle" (since that makes it dimensionless). So the solid angle is approximately $\pi \epsilon^2/r^2$ for small $\epsilon.$ (Real spheres, you don't see all the way to the equator if you are looking down at it from the pole, but you get arbitrarily close when your distance is much larger than the radius of the sphere.) 
We therefore find an equivalent perspective to Coulomb's law: First do this "smearing out" by $\epsilon$ and then take the limit as $\epsilon\to 0,$ the force of a point charge is some constant times the limit of $\epsilon$ times the smeared charge density $\rho_\epsilon$ times the solid angle $\alpha_\epsilon$ that the charge makes on the screen. The $1/r^2$ dependence is contained in this idea of "just look at the solid angle."
Now if you look at an infinite line of charge with charge density $\lambda,$ the above procedure performed on each charge separately will turn the whole thing into a little tube of diameter $2\epsilon$ and charge density $\rho_\epsilon = \lambda/(\pi\epsilon^2).$ The full force will have some multiplicative factor due to the fact that some components of the force are cancelling out, but this only depends on geometry; it is the same for all thin charged wires and not dependent on $\epsilon$. Nevertheless, we can immediately answer "if this were twice as far away, the factor of $\epsilon$ and the charge density $\rho_\epsilon$ would not change, but the solid angle would be half as much. Everything else would be the same because the line is infinite, only the solid angle would change." So in the limit, the effect of doubling your distance to the line of charge is to reduce the force by half. Therefore the force must be proportional to $\lambda / r.$ You still need calculus to get the multiplicative constant because the different forces from different directions are cancelling out; the only way I see to remove this step would be to somehow focus on the electric potential which adds like a scalar: but even then you need to calculate the slope of it, which sounds like a calculus problem.
Similarly, with the infinite sheet of charge, the above smears it out into a charge density of a plate with thickness $2 \epsilon$ and charge density $\sigma/(2 \epsilon).$ When you go twice as far away from it, the sheet has the same charge density and the same prefactor $\epsilon$ weights it, but what happens to the solid angle? Well if it's really an infinite sheet then the solid angle is still $2\pi$ since the solid angle of the whole sphere is $4\pi,$ and nothing has changed on the projected-sphere. So that's how you see directly that this effect of "there is more charge in a given amount of solid angle" has perfectly balanced out the effect of "that charge is further away and needs to be reduced by a larger constant when I project it onto the sphere." So the force must be independent of distance and proportional to $\sigma,$ again with some geometric factor. 
A: The answer to Electric field due to a line of charge explains how to use and interpret the Gauss' law result for an infinite line of charge.
When you consider an infinite plane, it does not matter how close you are to the plane. If you stand $1m$ away from the plane or a light year away from the plane, you won't be able to tell the difference. As you move away from the plane, you still see an infinite plane. You need to imagine this situation.
For the above reason, the electric field due to an infinite plane at a point $r$ units away from the plane is independent of $r$. It does not matter if the electric field varies as $\frac{1}{r}$ or as $\frac{1}{r^2}$. This is a unique property for an infinite plane. You can't tell how far you are from the plane.
