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.

 

[Meir Achuz]

I am confused by this thread. Perhaps I misread something, Why try to replace a trivial Gauss's law solution by an ambiguous limit of much more complicated wires?

 

[Vincf]

The Coulomb integral giving the horizontal field does not converge (its value depends on how far the surface is extended to infinity). This is why the application of Gauss's theorem is problematic: it says nothing about the horizontal field. Generally, when using this method, we say that "by symmetry," the horizontal component is zero. But the symmetry argument is not applicable here because the solution is not unique (for the horizontal field). The image of a uniform horizontal field under a plane of symmetry is another uniform horizontal field.

There is another clear example that illustrates the problem (so clear that it is never mentioned when applying Gauss's theorem): the field of an infinite uniform volume charge distribution. The entire space is filled with a uniform charge density. "By symmetry," we conclude that the field is zero, which is clearly false because it contradicts the local form of Gauss's theorem. In fact, the field of such a distribution depends critically on how the frontier goes to infinity and the solution is not at all unique.

 

[A.T.]

In fact, the field of such a distribution depends critically on how the frontier goes to infinity and the solution is not at all unique.

An infinite distribution doesn't have a "frontier" that goes somewhere. If you mean doing the integration asymmetrically, then you need a physical reason for that. And there is none in a symmetrical physical situation.

In physics, solutions which contradict physical properties of the scenario, are discarded.

 

[haruspex]

An infinite distribution doesn't have a "frontier" that goes somewhere. If you mean doing the integration asymmetrically, then you need a physical reason for that. And there is none in a symmetrical physical situation.

"An infinite mass is placed on each side of a beam balance. By symmetry, they balance."

In the real world, nothing is perfectly symmetrical. Idealisations are only valid if they work as the limit of a sequence of realistic versions; and if multiple limits are involved then it must not matter in which order you take them. School physics questions have been rightly lampooned as an irresistible force acting on an immovable mass.
There is a good reason for integrating asymmetrically: it is to test the validity of the symmetric treatment.

In physics, solutions which contradict physical properties of the scenario, are discarded.

An infinite plane of uniform charge is a non physical scenario. No appeal can be made to reality.

The standard textbook question has worth. It is clear to the student that it represents the case where the distance from the reference point to a uniformly charged plate is small compared with the distance to the perimeter of the plate.
If it were only to claim to find the normal field there would be no objection. The difficulty, as @Vincf has shown, is that the criterion for ignoring the tangential field is not that simple.

 

[A.T.]

In the real world, nothing is perfectly symmetrical... An infinite plane of uniform charge is a non physical scenario.

Sure, it's an idealization where the solution is simple. How good that approximation is, depends on the specific case, and what you care about.

I'm imagining a large, uniformly charged disk with σ = 1 C/m².
The disk's radius is 1 km. Point M is 1 mm above the disk and 250 m from its center. Therefore, the edges of the disk are between 750 m and 1250 m from the point. So, very far from the edges compared to the height above the plane. What is the electric field strength at point M?

The horizontal field is probably of the same order as the vertical field

Maybe I have missed it, but where is the calculation/estimation showing that the horizontal field is of the same order as the vertical field in the above scenario.

 

[Dale]

the solution is not at all unique

Which is what boundary values are for.

 

[Dale]

where is the calculation/estimation showing that the horizontal field is of the same order as the vertical field in the above scenario.

The horizontal field is arbitrary. It can be of any size, from negligible, to same order, to much larger. Although in this case the OP did not formally set it up as a boundary value problem the horizontal field is just a boundary value. There is also an additional constant vertical field that is neglected and which may also have any arbitrary value.

 

[haruspex]

Sure, it's an idealization where the solution is simple.

I'll say it again more strongly. The only reliable way to handle an idealisation is as the limit of realistic versions.
When there is more than one parameter to take to a limit, the relative rates of approach can affect the answer. In such cases, the problem is under-determined.
See e.g. the last few posts in https://www.physicsforums.com/threa...s-connected-by-a-spring.1081720/#post-7276500.

 

[A.T.]

Although in this case the OP did not formally set it up as a boundary value problem the horizontal field is just a boundary value.

Are you referring to a possible external electric field? I assumed that in @Vincf's scenario in post #39 this is implicitly zero. This should give a unique solution, and I wanted to know, if the horizontal component is indeed of the same order as the vertical field in that solution.

 

[Vincf]

We've already discussed this, and I don't share the view that it's a boundary value problem. Here, we're dealing with a simple summation problem: I'm given the imposed charges, and I'm looking for the field they create. For a finite charge distribution, the answer is given by the Coulomb integral. For an infinitely long wire, there's no problem: we simply extend the wire's terminals to infinity without worrying about how we extend them. For the vertical component of the plane, there's no problem: the vertical field is perfectly defined regardless of how the plane's edges tend towards infinity. For the horizontal component, it's different.

 

[PeroK]

We've already discussed this, and I don't share the view that it's a boundary value problem. Here, we're dealing with a simple summation problem: I'm given the imposed charges, and I'm looking for the field they create. For a finite charge distribution, the answer is given by the Coulomb integral. For an infinitely long wire, there's no problem: we simply extend the wire's terminals to infinity without worrying about how we extend them. For the vertical component of the plane, there's no problem: the vertical field is perfectly defined regardless of how the plane's edges tend towards infinity. For the horizontal component, it's different.

... because the horizontal component comprises two divergent integrals. Making the limit configuration fundamentally unphysical.

 

[Vincf]

We agree. The transition to the infinite limit is not "physical" for the horizontal component of the field. That is to say, with regard to the horizontal component, the "infinite plane" does not correctly model a finite distribution with dimensions much larger than the height. (More precisely: whose edges are far apart compared to the height). On the other hand, it models the vertical component of the same finite plane very well.

Note that this is not a "fundamental" problem! Its importance lies in the fact that the model is widely used in elementary electrostatics courses.
To summarize, this infinite plane is mainly used to find, in an elementary way, the field and capacitance of the parallel-plate capacitor.

Note that the infinite parallel-plate capacitor treated as a boundary value problem (with imposed potentials $V_1$ and $V_2$) does not pose the same kind of problem at all.

 

[Dale]

Are you referring to a possible external electric field?

With an infinite charge distribution I don’t think that “external” is meaningful. But, yes, if you are using this as an approximation for a finite distribution then it would be an external field.

Either way, it is a boundary condition.

I assumed that in @Vincf's scenario in post #39 this is implicitly zero.

To me, that is the interesting result of this thread. It is only implicitly set to zero if you use a specific symmetry. If you use a different symmetry then you are implicitly setting it to something else.

For Gauss’ law there is a constant of integration $\nabla \times \vec F$ where $\vec F$ is any smooth vector field. A specific $\vec F$ is chosen by the boundary conditions. So any term that crops up and can be written as $\nabla \times \vec F$ is absorbed in the boundary conditions. Regardless of whether it arises explicitly or implicitly. It doesn’t matter how it got there, all that matters is that it has the form of the constant of integration and is therefore the type of thing that is selected by the boundary conditions.

 

[A.T.]

But, yes, if you are using this as an approximation for a finite distribution then it would be an external field.
...
It is only implicitly set to zero if you use a specific symmetry. If you use a different symmetry then you are implicitly setting it to something else.

The scenario in post #39 is finite and not symmetrical. So there should be a unique non-zero solution for the horizontal component. My question was if the horizontal component is indeed of the same order as the vertical field for the numbers given in post #39, as @Vincf suggests.

 

[Dale]

The scenario in post #39 is finite and not symmetrical.

Ahh, I was speaking of the OP.

 

[Dale]

I don't share the view that it's a boundary value problem.

It is a mathematical fact that the terms that you obtain are included in the constant of integration. So, it is a mathematical fact that they are specified by setting the boundary value.

 

[Vincf]

It is a mathematical fact that the terms that you obtain are included in the constant of integration. So, it is a mathematical fact that they are specified by setting the boundary value.

I prefer to pose the question this way: I have a finite charged plane and a point $M$ above the charged plane at a height $z$ much smaller than the distance to the edge. Does an "infinite plane" model allow me to predict the field at $M$ without worrying about the details of the plane's shape?
For the vertical component, the answer is yes; for the horizontal component, it's no.
Since it's a finite plane, the boundary condition is $V = 0$ at infinity.

 

[Dale]

I prefer to pose the question this way

Sure, you can pose it many different ways. But they do not change the mathematical facts I mentioned above.

I have a finite charged plane and a point M above the charged plane at a height z much smaller than the distance to the edge. Does an "infinite plane" model allow me to predict the field at M without worrying about the details of the plane's shape?

Yes, provided that you specify reasonable boundary conditions.

the boundary condition is V=0 at infinity.

That isn't a reasonable boundary condition if you want to use the infinite plane solution for a finite plane of charge.

You can, in fact, take a finite charge plane, specify the boundary values on a boundary where the finite plane bisects the spatial boundary and there are no other charges in the boundary, and the solution for the infinite charged plane is valid. In other words, in that boundary, the finite charge plane and the infinite charge plane match, so the solution also matches.

If you move your boundary conditions out to infinity then a finite plane doesn't match an infinite plane inside the boundaries, so you should have no expectation that that would work.

For the vertical component, the answer is yes; for the horizontal component, it's no.

My answer is the same for the vertical and horizontal components.

 

[A.T.]

On page 12, figure 9, you have the curves of the normal and radial components of the field as a function of altitude. You'll see that the radial component is not at all negligible (for example, at 0.5R from the center, the radial componet is 1/3 of the z component).

Indeed, I ran a simple test, and got similar numbers.

You can also look at figure 11 and 12 which shows that in the case of the parallel plate capacitor, the radial components of the two faces cancel each other out in pairs so that the field between the two disks is significantly more uniform than the field created by a single disk.

Yes, that makes sense.