Horizontal component of the electric field of an infinite uniformly charged plane

[PeroK]

I interpret the standard result for the uniformly charged infinite plane as follows.

If we have a large uniformly charged plate, then the electric field above the centre is approximately constant for small enough distances. The distance depends on the size of the plate. Eventually, the field reduces to zero far enough from the plate.

For a hypothetical infinite "square" plate, the field is constant everywhere above the plate.

Although in a sense geometrically, an infinite Rectangular plate is the same shape as an infinite square plate, it is not the same. How the relative dimensions tend to infinity is important. The examples where apparently $\pi =4$ and $\sqrt 2 =2$ come from a similar lack of care in using limits.

More simply, two sequences $a_n, b_n$ might both tend to infinity, but we cannot conclude that the sequences are essentially equivalent from any other perspective. E.g. the function $e^x$ tends to infinity must faster than any fixed power of $x$. This is important. Any argument that vaguely assumed that $x$ and $e^x$ end up at the same $+\infty$ is potentially flawed. Formally, the functions are asymptocally very different.

The asymptotic behaviour of the dimensions for a rectangular plate is important. Hence, so is the order that we take them to infinity.

 

[haruspex]

That isn’t true. $\infty - \infty$ is an indeterminate form.

Right, which is why it is a fallacy to claim it equals zero.
It is an axiom that if x=a and x=b then a=b.

Did you mean that it's fallacy to claim that zero is a unique solution for the two-way infinite limit?

That is what would be implied by the statement that the limit equals zero.

 

[Dale]

Right, which is why it is a fallacy to claim it equals zero.

Why not? Finding an indeterminate form doesn’t prevent a limit from being equal to zero. That is more or less the whole idea behind derivatives.

And in this case all of the different limits are valid solutions to the problem.

 

[haruspex]

Why not? Finding an indeterminate form doesn’t prevent a limit from being equal to zero.

And in this case all of the different limits are valid solutions to the problem.

If we conclude $x=a$ (where a is well defined) then we necessarily mean that it is the only value of x that satisfies the conditions. This follows from the field axiom I quoted.
Given the condition $x^2=a^2$, we can conclude that $x\in\{a,-a\}$, not that $x=a$.
This is different from the question of whether a “ is a solution “.

 

[Dale]

If we conclude $x=a$ (where a is well defined) then we necessarily mean that it is the only value of x that satisfies the conditions. This follows from the field axiom I quoted.
Given the condition $x^2=a^2$, we can conclude that $x\in\{a,-a\}$, not that $x=a$.
This is different from the question of whether a “ is a solution “.

Indeed.

 

[Vincf]

For the case of a finite plane, which is the most interesting, we could extend the use of Gauss's theorem with a slight modification. We take the traditional Gaussian surface, for example, a vertical cylinder extending from $z$ to $-z$ but with a small radius $r$. The flux of the field through this cylinder is the sum of the fluxes at the top and bottom, $2 E_z \pi r^2$, and the lateral flux.

To find this lateral flux, we note that what matters is the variation of the horizontal field over the length $r$. Let $D$ be the typical length of variation of the lateral field. We can assume that $D$ is on the order of the minimum distance to the edges.

If we take $\sigma\ / \epsilon_0$ for the typical dimension of the horizontal field, its derivative will typically be $(\sigma\ / \epsilon_0) (1/D)$ and therefore the variation of the horizontal field over the length $r$ will be $(\sigma\ / \epsilon_0) (r/D)$. The corresponding flux will be on the order of this variation multiplied by the lateral surface $2 \pi r z$, that is $(\sigma\ / \epsilon_0) r^2 (z/D)$.

In total, Gauss's theorem could be written as: $$2 E_z \pi r^2 + (\sigma\ / \epsilon_0) r^2 (z/D) = (\sigma\ / \epsilon_0) \pi r^2 $$ We indeed find the value $E_z = \sigma\ /( 2 \epsilon_0)$ with a relative error on the order of $(z/D)$ and we can say nothing about the horizontal component.

If, in addition, we add the existence of a "center," that is, symmetries allowing us to conclude that the horizontal field is zero, we find the infinite plane studied in our courses.

For the case of the mathematical "infinite plane." It's a convergence problem. It involves finding the limit of a sum as the charge distribution is extended indefinitely. So it's not a boundary condition problem. The integral giving the horizontal component of the field is a double integral of a function that is not of constant sign and does not converge. The only mathematics books I own that address this subject are French. There is a seemingly well-known example of a non-convergent integral: the Cayley integral. I found this link in French :
http://serge.mehl.free.fr/chrono/Cayley.html
If we calculate it in Cartesian coordinates :
$$
\iint_{D} \sin(x^2+y^2)\, dx\, dy
$$
and take the limit of the infinite plane, we find π/4,
But if we switch to polar coordinates :
$$
\iint_{D'} \sin(r^2)\, r\, dr\, d\theta
$$
we find that the integral is divergent (has no limit)!

 

[PeroK]

Such paradoxes of divergent series are well known. For example:

Take any divergent series $a_n$ and any real number $L$, then by appropriate manipulation of the order of terms, it can be shown that:
$$\sum_{n=1}^{\infty}a_n = L$$

 

[PeroK]

In general, in real analysis, you have to show first that a series is convergent before you can prove that it converges to a real number.

You are simply doing a physicists non rigorous mathematics and being surprised when divergence bites you back!

 

[Vincf]

Hello,
Indeed, for series that are not absolutely convergent, the problem is well known. And when I was a student, in France, the math professor didn't joke about it! The physics professor was much less strict. But as physicists, we have to be careful from time to time.
There's the wonderful "Riemann rearrangement theorem" which states that:
https://en.wikipedia.org/wiki/Riemann_series_theorem
"if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, and rearranged such that the new series diverges."

While writing this post, I vaguely remembered a remark from my chemistry professor about convergence problems in obtaining the Madelung constant (alternating series). And indeed, the problem is mentioned on the Wikipedia page:
https://en.wikipedia.org/wiki/Madelung_constant
With an article of which I only read the summary:
https://pubs.aip.org/aip/jmp/articl...e-sums-and-Madelung-s?redirectedFrom=fulltext

 

[Dale]

The integral giving the horizontal component of the field is a double integral of a function that is not of constant sign and does not converge.

In post 1 you demonstrated multiple ways to take the limit that do converge. It does seem to be conditionally convergent. And each of the fields to which it conditionally converges are valid solutions to the original problem.

 

[PeroK]

In post 1 you demonstrated multiple ways to take the limit that do converge. It does seem to be conditionally convergent. And each of the fields to which it conditionally converges are valid solutions to the original problem.

The integral is not absolutely convergent. It's divergent improper integrals that cancel each other out by symmetry.

If there is no symmetry, then the result is not well defined, as shown in the OP. And, as one would expect.