# Significance of electrical fields of infinite objects

I've noticed a pattern with the electric fields of charged objects of infinite dimensions.

A point charge, which can be considered a charge of 0 dimensions, has an electric field that goes as $r^{-2}$. An infinite line of charge, a 1 dimensional object, has an electric field that goes as $r^{-1}$. The infinite plane (2D) has a constant electric field, or a field proportional to $r^0$. And if one considers a sphere of electrical charge, you can make a Gaussian sphere inside it, and this electric field should grow as $r^1$.

I'm trying to figure out an explanation for this pattern. The original question that led me to consider this was the difference between the infinite line and the infinite plane: why does one depend on r but not the other? It's addressed briefly in Griffiths E&M, where I believe he says something along the lines of "you can't get away from an infinite plane". But by that reasoning, I don't understand how you can "get away" from an infinite line, either. The reasoning make some sense if you consider an infinite plank, a plane that is infinite in one dimension and finite in another, so that as you get further away, one dimension "shrinks" while the other continues to "appear the same". However, the "perfect" infinite line should only have one dimension to begin with, so it should "appear the same" no matter how far away or close you are...

I suppose my questions are these: why does this pattern between the dimensions of an object and its field exist? How do you "get away" from an infinite line?

I believe the answer to your question lies in the Gauss theorem itself $$\oint \textbf{E}d\textbf{S} \sim Q$$ and the symmetry of the system, which defines the shape of equipotential surfaces.

1. In case of a point charge there is a rotational symmetry about any axis going through the charge, so the equipotential surfaces are spheres whose area is proportional to $r^2$. Thus the field is $\sim r^{-2}$ ($r$ is distance from the charge)

2. In case of a line, rotational symmetry exists only around this line, so equipotential surfaces are cylinders, whose area (per length along the line) is $\sim r$ (here $r$ is the shortest distance to the line). Thus the field is $\sim r^{-1}$

3. For a plane, equipotential surfaces are planes, parallel to the given plane, their area (again, per unit area of the given plane) is independent of the distance from the given plane, as is the field.

4. The case of sphere is not like the ones above! Here you are inside the sphere, so the charge $Q$ on the rhs of gauss law also changes with the distance from the center, proportionally to the volume $\sim r^3$. Equipotential surfaces are spheres again, whose area is $\sim r^2$, so the field increases linearly with $r$. If you were outside the sphere the field would decrease $\sim r^{-2}$ as in case 1.

I don't like the "you can't get away" explanation. There is a simple explanation with field lines:

In all three cases, the field lines are straight lines from the point charge to infinity. You can easily calculate the density of the field lines for each object. For a point charge, the "number" of field lines through any sphere around the point charge is the same (as all field lines are straight, they have to pass any sphere centered at the charge exactly once). The density is $n/A$, where $n$ is the number of lines and $A$ is the surface area of the sphere. Therefore, since the number of lines is constant and the surface of the sphere is $r^2$, the density goes as $1/r^2$.

For a line, the density only thins out around circles of the line. In other words, the number of field lines through each cylinder around the line is the same. Since the area of the cylinder is proportional to $r$, the density of field lines grows as $1/r$.

Finally, for the plane, the number of lines through any plane parallel to the plane is the same. Since the surface doesn't change, the density is constant (this is also true, because all field lines are parallel).

The only question that remains is why the field strength is proportional to the density of the field lines. Why does that work? I guess it's somehow intuitive, but for a technical answer let me refer to the excellent answer by Emilio Pisanty here: Why does the density of electric field lines make sense, if there is a field line through every point?