# Explanation for $E~$ not falling off at $1/r^2$ for infinite line and sheet charges?

For an infinite line charge, $E$ falls off with $1/r$; for an infinite sheet of charge it's independent of r! The infinitesimal contributions to $E$ fall off with $1/r^2$, so why doesn't the total $E$ fall off the same way for the infinite line and sheet charges?

Loosely speaking, as we walk away from a sphere it looks smaller, as we walk away from a cylinder just the radius looks smaller, but not the infinite length, and finally as we walk away from an infinite sheet of charge it never looks any smaller (we can never 'get away' from an infinite sheet).

At more mathematical level I would say the best way to see this is with Gauss's Law. I am going ignore the details below and just assume you have a clue what Gauss's Law is, otherwise the rest of the post is pretty irrelevant for you. Below I will use the following 3 Gaussian surfaces I have borrowed from

For any case where you are interested in obtaining the electric field using Gauss's Law $\displaystyle \frac{ q_{enc}}{\epsilon_o} = \Phi= \int \bf{dA} \cdot {E} ~$ you will end up picking a gaussian surface such that the electric field is constant or zero over the Gaussian surface so that it just evaluates to $|\bf{E}| \int |d\bf{A}_{||}|$ where the surviving integral is just over portions of the area parallel to the electric field. For example, for a point charge you will draw a concentric sphere. You can see from the above the further away you get from the point charge the bigger the area gets and so the electric field must get smaller in order to keep the flux constant. For a line charge we would draw a concentric cylinder and only the portion of the area that wraps around will survive. Again the flux must be constant but the area that wraps around a cylinder that grows much more slowly as the radius grows than that of the surface area of a sphere as the sphere's radius grows. Finally for a sheet we would draw a cylinder so that only the ends of the cylinder that are parallel with the sheet survive. Now as the length of the cylinder increases the surface area of the ends doesn't grow at all, so the electric field doesn't fall off at all.

The true explanation is, of course, the math. But I'm guessing you're familiar with the calculation. For an intuitive understanding, I'd put it like this: when you're very close to an infinite sheet of charge, the contributions to $\vec{E}$ of the pieces of charge far away from you mostly cancel out, because e.g. the electric field of a bit of charge far to your left is nearly antiparallel to the field of a bit far to your right. When you're further away from the sheet, then those contributions to the field are not so close to antiparallel, and they add up to a larger net electric field. That balances out the reduced electric field from the bits of charge closest to you.

• +1: When I'm teaching this to students, this type of answer usually seems to satisfy them the most Jul 27, 2012 at 2:30

$E$ falls off at $1/r^2$ when there are 3 degrees of freedom for the field lines to spread. When you have an infinite line in three dimensional space, that's equivalent to having the field spread from a point in a cross section of this space, which goes as $1/r$. Similarly, field lines spreading from a flat sheet have only 1 remaining degree of freedom to spread in, it's like spreading from a point in 1-dimensional space, which goes as $1/r^0$.

• I don't understand your answer at all, but thanks for the effort. Jul 8, 2013 at 15:15
• @Physikslover: What exactly did you not understand ? . (You do know that the $r^2$ term comes from the surface area of a sphere , right?) . Aug 13, 2013 at 12:53
• Your argument is correct, but the intuition is from field line spreading, so the better way to say it (in such a way that someone else can also understand) is to use Gauss's law, this is the field line counting made clear. Aug 22, 2013 at 22:09
• @RonMaimon now that you've had a long rest, does this answer still make sense to you honestly? Aug 25, 2014 at 2:26