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

[Vincf]

TL;DR

Convergence problems for integrals giving the horizontal component of the field of an infinite plane

Hello,

I would like to discuss a mathematical point concerning the electric field of an infinite uniformly charged plane.
This question arose while I was discussing an electrostatics exercise with a student. I wrote an article about this (in 2023) in a French journal for french physics professors. I made it available online as a postprint at this link: https://zenodo.org/records/18431824
(But I continued to teach the usual result since it was in the official curriculum for my students !)
The usual result is that the field is strictly perpendicular to the plane and has magnitude $E = \frac{\sigma}{2\varepsilon_0}$
Below, I reproduce what is a classic electrostatics exercise, which presents the problem quite clearly.
I compute the electric field of an infinite uniformly charged plane by decomposing it into infinite parallel wires.
I redrew a figure to make it easier to read :

Each wire, parallel to the x-axis, has linear charge density : $d\lambda = \sigma \, dy_P$ and produces the field :
$$
d\vec{E} =
\frac{d\lambda}{2\pi\varepsilon_0}
\frac{1}{PM}
\, \vec{PM}
$$
where $r=PM$ is the distance from the wire to the observation point $M$.
Projecting onto the coordinate axes gives : (The notations are on the diagram.)
$$
dE_y =
\frac{\sigma \, dy_P}{2\pi\varepsilon_0}
\frac{1}{PM}
\cos(\alpha)
$$
$$dE_z =\frac{\sigma \, dy_P}{2\pi\varepsilon_0}\frac{1}{PM}\sin(\alpha)$$
From geometry, $PM^2=r^2 = (y_M - y_P)^2 + z_M^2$ and $(y_M - y_P) dy_P = -r \, dr$ so $\frac{dy_P \cos(\alpha)}{r}=-\frac{dr}{r}$
Also,
$\tan(\alpha) =\frac{z_M}{y_M - y_P}$ or $-dy_P =z_M \, d\left(\frac{1}{\tan(\alpha)}\right)$ so $dy_P = \frac{z_M}{\sin^2\alpha} \, d\alpha$ and $\frac{dy_P \sin(\alpha)}{r} =d\alpha$
Thus,
$$
dE_y =- \frac{\sigma}{2\pi\varepsilon_0}\frac{dr}{r}
$$
$$
dE_z =\frac{\sigma}{2\pi\varepsilon_0}\, d\alpha$$
Integrating :
$$E_y =- \frac{\sigma}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)$$
$$

E_z =

\frac{\sigma}{2\pi\varepsilon_0}

(\alpha_{\max} - \alpha_{\min})
$$
It remains to pass to the limit of the infinite plane. $D_{\max} \to \infty, \quad D_{\min} \to \infty$
For the vertical component, $\alpha_{\max} \to \pi, \quad \alpha_{\min} \to 0$ gives $E_z = \frac{\sigma}{2\varepsilon_0}$ as expected.
However, for the horizontal component,
$$

\lim_{D_{\max}\to\infty \atop D_{\min}\to\infty}

E_y

=

- \frac{\sigma}{2\pi\varepsilon_0}

\lim_{D_{\max}\to\infty \atop D_{\min}\to\infty}

\ln

\left(

\frac{

\sqrt{(D_{\max} - y_M)^2 + z_M^2}

}{

\sqrt{(D_{\min} + y_M)^2 + z_M^2}

}

\right)
$$
The result appears to depend on how the limit is taken.
If $D_{\max} = D_{\min}$ then :
$$

\lim_{D\to\infty \atop D_{\max}=D_{\min}=D}

E_y = 0

$$
But if $D_{\max} = K D_{\min}$ (With $K$ a positive constante):
$$
\lim_{D_{\min}\to\infty \atop D_{\max}=KD_{\min}}
E_y =
- \frac{\sigma}{2\pi\varepsilon_0}\ln(K)
$$
This shows that the horizontal component obtained from direct integration depends on the procedure used to take the infinite limit.

My question is the following:
When computing the field of an infinite uniformly charged plane using this decomposition into parallel infinite wires, what is the mathematically rigorous procedure for taking the infinite limit?
In particular, why is it justified to interpret the integral as a principal value, corresponding to a symmetric extension of the plane around the observation point M?

 

[Dale]

Do you have something against symmetry? I mean, is there a specific reason that you are going out of your way to avoid using symmetry?

 

[PeroK]

My question is the following:
When computing the field of an infinite uniformly charged plane using this decomposition into parallel infinite wires, what is the mathematically rigorous procedure for taking the infinite limit?

If I have time, I'll look at the details of what you've done. In general, if you have two limits, then the order can make a difference. Let's assume we have two sequences $a_n = 2^n, b_n = \frac{1}{2^{2n}}$.

It's clear that $a_nb_n = \frac 1 {2^n}$ and:
$$\sum_{n=1}^{\infty} a_nb_n = \sum_{n=1}^{\infty} \frac 1 {2^n} = 1$$However, the following construction diverges:
$$\bigg (\sum_{n=1}^{\infty} a_n \bigg )\bigg (\sum_{k=1}^{\infty} b_k \bigg )$$
PS maybe this isn't the best example, but hopefully it makes the point that having one limit is clear; whereas, have two limits is potentially ambiguous.

In particular, why is it justified to interpret the integral as a principal value, corresponding to a symmetric extension of the plane around the observation point M?

If we are modelling the physical scenario where we have a large square plate of uniform charge, and consider a point above the centre of the plate, then that should be the configuration that should be taken to the limit directly, in order to get a calculation in the limit where the plate is very large.

It might work if you took a different configuration (a rectangle or a circular disk) and increased the dimensions. The problem in the case of the rectangle is that you have two dimensions to vary. You'd need to be careful there. Taking the limit of a circular disk should work just as well as a square.

In the case of modelling the plate as a set of long wires, you could have $n$ wires of length $l$, separated by a distance $d$ and a uniform linear charge density $\lambda$. However, you would be best to have one variable that determines all these variables so that they approximate a square of charge of approximately constant surface charge density. [PS although I'm not sure how that would work. You'd have to be very careful.]

Starting with infinite wires looks suspicious, as the configuration is never symmetric in the two planar directions. I'm not surprised if that goes wrong.

 

[Vincf]

Hello and thank you for your reply!

No, I have nothing against symmetries. But it seems difficult to impose a priori on the concept of an "infinite plane" the requirement of symmetry. For the concept to be useful, it must model a real-world situation: a charged sheet of finite dimensions at a point that is at a very small height compared to the distance to the edges. For example, when I deduce the field of a parallel-plate capacitor from the field of two infinite planes, I don't say that one must be at the center of the parallel-plate capacitor.
When we talk about an "infinite wire," we don't need to add that it is symmetrical. It is sufficient that the ends of the wire are far apart compared to the distance to the wire. And in the case of the wire, we can make the ends tend towards infinity in any way without any problem.

Furthermore, I believe the symmetry argument is problematic here because the solution for the horizontal component is not unique. The image of a uniform horizontal field with respect to a plane of symmetry is another uniform horizontal field. The equation $x^2=1$ is invariant if we change $x$ to $-x$, but we don't conclude that the solution is 0. When we use a symmetry argument, we say "the field must be in this direction," and if we mirror the system, it must be in the opposite direction: therefore, it is zero. This is only true if the solution is unique.

 

[Vincf]

If I have time, I'll look at the details of what you've done. In general, if you have two limits, then the order can make a difference. Let's assume we have two sequences $a_n = 2^n, b_n = \frac{1}{2^{2n}}$.

It's clear that $a_nb_n = \frac 1 {2^n}$ and:
$$\sum_{n=1}^{\infty} a_nb_n = \sum_{n=1}^{\infty} \frac 1 {2^n} = 1$$However, the following construction diverges:
$$\bigg (\sum_{n=1}^{\infty} a_n \bigg )\bigg (\sum_{k=1}^{\infty} b_k \bigg )$$
PS maybe this isn't the best example, but hopefully it makes the point that having one limit is clear; whereas, have two limits is potentially ambiguous.

If we are modelling the physical scenario where we have a large square plate of uniform charge, and consider a point above the centre of the plate, then that should be the configuration that should be taken to the limit directly, in order to get a calculation in the limit where the plate is very large.

It might work if you took a different configuration (a rectangle or a circular disk) and increased the dimensions. The problem in the case of the rectangle is that you have two dimensions to vary. You'd need to be careful there. Taking the limit of a circular disk should work just as well as a square.

In the case of modelling the plate as a set of long wires, you could have $n$ wires of length $l$, separated by a distance $d$ and a uniform linear charge density $\lambda$. However, you would be best to have one variable that determines all these variables so that they approximate a square of charge of approximately constant surface charge density. [PS although I'm not sure how that would work. You'd have to be very careful.]

Starting with infinite wires looks suspicious, as the configuration is never symmetric in the two planar directions. I'm not surprised if that goes wrong.

Hello and thank you for your reply!
Actually, the infinite-wires decomposition is a way to decompose the plane to calculate the integral over the charged surface. We could avoid using the term "infinite wires": write the double integral, integrate over $x$, then over $y$. This is what we always do when calculating a double integral. We could also integrate first over $y$ and then over $x$, and there we would find 0. What puzzles me is that if an integral of this type converges, then the result shouldn't depend on how it's calculated. The integral shouldn't depend on how we make the edges tend towards infinity.

This is what happens for the integral of the vertical component: its sign is constant, and we get a finite result with a decomposition, so we can be sure that it converges.

Whereas for the horizontal component, we only find 0 if we impose a certain way of making the edges tend towards infinity. And that seems to me to be the sign of an integral that does not converge.

 

[PeroK]

No, I have nothing against symmetries. But it seems difficult to impose a priori on the concept of an "infinite plane" the requirement of symmetry.

An infinite uniformly charged plate is a physically impossible construction in classical electromagnetism. A charged particle has infinite potential energy in the electric field.

In any physically realisable scenario, the distance from the plate must eventually be significant compared to the size of the plate.

In general, an infinite amount of charge is always potentially problematic. It's important to understand the limitations of taking an infinite integral. This is always an approximation to a finite sum.

 

[PeroK]

What puzzles me is that if an integral of this type converges, then the result shouldn't depend on how it's calculated. The integral shouldn't depend on how we make the edges tend towards infinity.

If you have have linear wires, then the linear charge density must tend to zero, as the number of wires increases. You end up with effectively an infinite number of wires of zero charge. You must take the linear charge density to zero at the same time as taking the number of wires to infinity. And, you must keep the effective surface charge constant.

You cannot take the two limits in any order. Taking one limit first will tend to give you an answer of zero.

 

[Dale]

But it seems difficult to impose a priori on the concept of an "infinite plane" the requirement of symmetry.

Why? It seems very clear that for every point “up and to the left” there is a symmetrical point “down and to the right”. Using that fact immediately tells you that there is no E field parallel to the plane.

 

[Vincf]

I think we could forget about infinite wires and the diagram and work purely mathematically. We need to calculate a limit of an integral:
We wish to compute the generalized integral:

$$
I_x =
\int_{-\infty}^{+\infty}
\int_{-\infty}^{+\infty}
\frac{x_M - x_P}
{\left((x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2\right)^{3/2}}
\ dx_P \ dy_P
$$
Classically, in Cartesian coordinates, there are two ways to compute this integral. If this integral converges, its value does not depend on how the limit is taken and therefore, in particular, on the order of integration in Cartesian coordinates.
The first method consists in writing:
$$
I_x =
\int_{-\infty}^{+\infty} dy_P
\int_{-\infty}^{+\infty}
\frac{x_M - x_P}
{\left((x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2\right)^{3/2}}
\ dx_P
$$
The inner integral exists and can be written:
$$
\int_{-\infty}^{+\infty}
\frac{x_M - x_P}
{\left((x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2\right)^{3/2}}
\ dx_P
=
\left[
\frac{-1}
{\sqrt{(x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2}}
\right]_{x_P=-\infty}^{x_P=+\infty}
=
0
$$
This allows us to conclude:
$$
I_x =
\int_{-\infty}^{+\infty} 0 \, dy_P = 0
$$
However, one may proceed in the reverse order:
$$
I_x =
\int_{-\infty}^{+\infty}
(x_M - x_P)
\left(
\int_{-\infty}^{+\infty}
\frac{1}
{\left((x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2\right)^{3/2}}
\, dy_P
\right)
dx_P
$$
Again, the inner integral exists and can be written:
$$
\int_{-\infty}^{+\infty}
\frac{1}
{\left((x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2\right)^{3/2}}
\ dy_P
=
\left[
\frac{y_P - y_M}
{((x_M - x_P)^2 + z_M^2)
\sqrt{(x_M - x_P)^2 + (y_M - y_P)^2 + z_M^2}}
\right]_{y_P=-\infty}^{y_P=+\infty}
$$

which gives:

$$
=
\frac{2}
{(x_M - x_P)^2 + z_M^2}
$$
It then remains to compute:
$$
I_x =
\int_{-\infty}^{+\infty}
\frac{2(x_M - x_P)}
{(x_M - x_P)^2 + z_M^2}
\ dx_P
=
\left[
-\ln\left((x_M - x_P)^2 + z_M^2\right)
\right]_{x_P=-\infty}^{x_P=+\infty}
$$

But this time, the integral does not exist.

The reasoning is identical for the component along the y direction.

 

[Dale]

I think we could forget about infinite wires

Yes. I wouldn’t use wires at all. I would use rings. It simplifies substantially and requires only one limit.

As mentioned by others, taking multiple limits is inherently ambiguous. So I try to avoid it if possible.

 

[Vincf]

Why? It seems very clear that for every point “up and to the left” there is a symmetrical point “down and to the right”. Using that fact immediately tells you that there is no E field parallel to the plane.

Why? It seems very clear that for every point “up and to the left” there is a symmetrical point “down and to the right”. Using that fact immediately tells you that there is no E field parallel to the plane.

It seems to me that this reasoning would be valid for a finite distribution. But for an infinite distribution, it doesn't seem so simple. An integral can converge in principal value without necessarily being convergent.

We encounter this problem awith the "infinite volume distribution": the entire space filled by a uniform distribution of charges. Then, we are tempted to conclude that, "by symmetry," the field is zero everywhere. To each charge $dQ$ on one side, we can associate a symmetric charge that cancels the field. However, this result is clearly in contradiction with the Maxwell-Gauss equation: zero field and non-zero charge density. The problem arises from the fact that the limit is poorly defined: the field depends critically on the precise shape of the distribution at a distance.

 

[Dale]

It seems to me that this reasoning would be valid for a finite distribution. But for an infinite distribution, it doesn't seem so simple.

Can you identify any point in an infinite plane where there is a point "up and to the left" but not a symmetrical point "down and to the right"?

We encounter this problem awith the "infinite volume distribution": the entire space filled by a uniform distribution of charges. Then, we are tempted to conclude that, "by symmetry," the field is zero everywhere. To each charge dQ on one side, we can associate a symmetric charge that cancels the field. However, this result is clearly in contradiction with the Maxwell-Gauss equation: zero field and non-zero charge density. The problem arises from the fact that the limit is poorly defined: the field depends critically on the precise shape of the distribution at a distance.

This is a good point.

What you can always do is go back to Maxwell's equations to find if other solutions are consistent with the equations. I have not done that with your additional solutions, so perhaps they are valid.

 

[Vincf]

Yes. I wouldn’t use wires at all. I would use rings. It simplifies substantially and requires only one limit.

As mentioned by others, taking multiple limits is inherently ambiguous. So I try to avoid it if possible.

But when we write the integrals that give the electrostatic field, we don't have to specify the partitioning associated with the integral. Or does the integral lose its meaning?

I haven't done the calculation, but I think that, for example, if we consider an elliptical charge distribution with point $M$ vertically above the focus of the ellipse (and not the center), and then make the ellipse tend towards infinity while keeping its eccentricity constant (i.e., a single parameter), we wouldn't find 0 for the horizontal component of the field. Whereas the vertical component would indeed tend towards the standard value.
By imposing a circular partitioning, I think you're imposing a circular symmetry on the "infinite plane," which shouldn't be part of its definition ?

 

[PeroK]

You should be able to use any shape (except mathematically pathological ones) and the integral convergence should work.

The infinite plane is geometrically an uncountable collection of lines. But, standard calculus is going to struggle with that. Because the area of any line is zero.

An countable infinite collection of strips of finite width will work.

Best not to mix line integrals, surface integrals and volume integrals in this context.

 

[A.T.]

I think you're imposing a circular symmetry on the "infinite plane,"

What asymmetry could there be on the infinite plane? Isn't it symmetrical around any point it contains?

 

[Baluncore]

I think there is a contradiction, hidden in the assumptions.

If the charged plane is infinite in extent, then there must be a boundary condition surrounding the space, such as another plane of infinite extent, at zero potential, but at an infinite distance.

Now consider the capacitance between the planes. The infinite extent of the planes gives infinite capacitance, while infinite separation of the planes will give zero capacitance. There is a fundamental contradiction.

 

[Dale]

I think you're imposing a circular symmetry on the "infinite plane," which shouldn't be part of its definition ?

Again, why not? It does have circular symmetry. If you do not consider that justified, then on what grounds are you justified in imposing a linear symmetry by using infinite lines of charge?

What you can always do is go back to Maxwell's equations to find if other solutions are consistent with the equations. I have not done that with your additional solutions, so perhaps they are valid.

So looking at Maxwell's equations, it does seem that a uniform E field with no source is a valid solution to Maxwell's equations. And since they are linear, you can add any uniform E field to the "usual" solution and get another valid solution. This is essentially a constant of integration that we usually take to be 0, but you are taking to be something non-zero.