# Scaling of Static Electric Field

The electric field of a point charge goes like $\displaystyle\frac{1}{r^2}$

The electric field of an infinite line goes like $\displaystyle\frac{1}{s}$

The electric field of an infinite plane is constant

Is there an intuitive way or some insight to understand such results and realize that they are expected?(One of my friends said that the electric field due to an infinite plane does not change at all because it looks infinite no matter how close or far from we are it. Is this argument correct? I am not convinced because we can say the same about the infinite line, it looks infinite no matter how close or far we are from it, yet it changes parametrically with the distance)

Gauss theorem in integral form gives rather an intuitive answer. Flux of electric field through a surface $D$ $\Phi=\int_D \vec E\cdot \mathrm{d} \vec A$ is proportional to the total charge $Q$, contained inside the surface. Now, compare three cases:
1) Point source. Draw spherical shells centered on the point source. Due to spherical symmetry $\vec E$ is normal to the shells and depends only on the radius of a shell. Hence $Q\sim \Phi=\int_D \vec E\cdot \mathrm{d} \vec A= E \int_D \mathrm{d} A=E A$. As the area of spheres goes as $A\sim r^2$, one has $E\sim A^{-1}\sim r^{-2}$.
2) Line source. Draw cylinders aligned with the source. Here the symmetry of the problem tells us that $\vec E$ is normal to the cylinders, and again depends only on the radius of a cylinder. Derivation is the same as before, except for that the area of a cylinder $A\sim r$ instead of $r^2$. Hence $E\sim r^{-1}$.
3) In the case of a plane, draw prisms with the base parallel to the plane and having the same area, so that each prism crosses the plane. Due to symmetry reasons the field is normal to the plane, and hence to the bases of the prysms. As the area of the base is the same for all the prisms, the charge inside the prisms is also the same. On the other hand $A$ doesn't depend on the height of the prisms at all. Hence $E\sim r^0$.