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

[PeroK]

Suppose we have a sequence of shapes that converge in some geometric sense to a limit shape - bounded or unbounded. That does not imply that the limit shape has properties that are the limit of the sequence of shapes. A common example is disproving Pythagoras theorem using an infinite sequence of staircase shapes of smaller and smaller step size.

The infinite plane is the limit of any number of different shapes. For example, a circle has a centre and an ellipse has two foci. It makes no sense to talk about whether the infinite planr has a centre or two foci. It's neither an infinite circle nor an infinite ellipse.

The same applies to infinite integrals. There is no guarantee that generating an infinite line or an infinite plane in two different ways will have the same convergence properties for a sequence of integrals.

Saying that it's the same line or plane in the limit is not a valid argument.

 

[Vincf]

Yes, I agree with you, a limit process can be problematic. But in physics, we use these kinds of infinite limits to approximate real-world situations. Therefore, the limit is only useful if it's close to the finite situation.

What I think is that this isn't the case for the infinite plane with respect to the horizontal component of the field. When we use the concept of the infinite plane as if it were a large finite plane, we can make significant errors.
We should, in another post, revisit the cases where, in our physics courses, we use this concept to model something. These cases don't seem that numerous to me.

Another problem this raises is the use of Gauss's theorem for limits whose behavior we don't know a priori. In the case of the infinite wire, everything is fine. But it's a more or less direct calculation that proves that. Indeed, the contribution of the distant parts of the wire isn't significant. And we use Gauss's theorem precisely to avoid direct calculation!

Ultimately, I get the impression that the application of Gauss's theorem is mainly clear for the spherical case. That's already quite something. Newton would have been happy to have this theorem !

 

[Dale]

There is a mathematical convergence problem for the horizontal component of the field.

It isn’t a convergence problem. None of the integrals or limits involved diverge.

 

[Dale]

A common example is disproving Pythagoras theorem using an infinite sequence of staircase shapes of smaller and smaller step size

That sounds very interesting. Do you have a reference handy for that?

 

[Dale]

What I think is that this isn't the case for the infinite plane with respect to the horizontal component of the field. When we use the concept of the infinite plane as if it were a large finite plane, we can make significant errors.

It isn’t an error for a person to find a valid solution to a problem. It isn’t an error for a different person to find a different valid solution to the same problem.

You have found a valid solution to the problem. The fact that your approach gives a different result is not a convergence issue nor an error.

Also, the vertical component has the same issue.

 

[Vincf]

By "error" I just meant that by using the model, we could predict the horizontal field to be zero, whereas in reality, it's of the same order of magnitude as the vertical field.
To be more precise, we should look at a concrete example of using the infinite plane model to see what it yields. Perhaps in another post so this one doesn't become too long?

As for the term "divergent": it's simply a matter of terminology. I call an integral that doesn't converge in the usual sense "divergent." So, for example, I would call an integral that converges only in its principal value a divergent. But it's just a matter of terminology.

 

[A.T.]

That sounds very interesting. Do you have a reference handy for that?

I think it similar to the proof that pi=4 in the video below at 1:39. The explanation at 15:09 points out the fallacy @PeroK mentions above.

 

[Dale]

To be more precise, we should look at a concrete example of using the infinite plane model to see what it yields. Perhaps in another post so this one doesn't become too long?

IMO already having 3 is confusing and fragmented. Besides, we have already discussed this: a plate capacitor close to the center of the plate.

By "error" I just meant that by using the model, we could predict the horizontal field to be zero, whereas in reality, it's of the same order of magnitude as the vertical field.

That is true for any boundary value problem. If you pick the wrong solution from the family of solutions then your solution will not be a good approximation to reality.

 

[Vincf]

The capcitor problem is different because we have two charged planes with opposite charges. The horizontal components cancel each other out.
But consider this question :
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?
Can I use the standard model of an infinitely charged, uniformly charged plane in this case?
If not, how do I define the conditions for the model to be usable?
If you tell me that I have to compare the distance to the center to the disk's radius, and that I have to stay close to the center, then it seems to me that the model loses its relevance. When discussing the infinite plane using Gauss's theorem, we don't talk about a "center" or a point of symmetry.

Another point: when discussing a parallel-plate capacitor, we don't stay close to the center of the disk to obtain a uniform field. We say the field is uniform as long as we are at a distance from the edges much greater than the distance between the planes: very far from the edge.

 

[Dale]

The capcitor problem is different because we have two charged planes with opposite charges.

It actually isn’t different if you are close to the center of one of the plates. And it is a good example to show that the “vertical” component also can have an arbitrary constant added too.

What is the electric field strength at point M?
Can I use the standard model of an infinitely charged, uniformly charged plane in this case?
If not, how do I define the conditions for the model to be usable?

To define the conditions for any model to be usable you have to specify how large an error you can accept for something to be considered usable. At point M one of the family will be exact. Near M, it will be an approximation. How near is near and how approximate is approximate depending on on the application

 

[Vincf]

The supposedly applicable conditions (see the quote from Walter Lewin) are that the height must be small compared to the distance from the edge. These conditions are correct for an infinite wire. They are correct for an infinite plane with respect to the vertical component, which, in the example I give, would be σ/2 epsilon 0 with a ridiculously small error.

But these conditions would be false for the horizontal component. The horizontal field is probably of the same order as the vertical field. Therefore, a plane with a radius of 1 km viewed from a height of 1 mm cannot be treated as "infinite," even if I am very far from the edges with respect to the horizontal component.

Yet, in electrostatics textbooks, the uniformity of the field between the plates of a parallel-plate capacitor is justified using the "zero horizontal component" model. I don't believe there is an undergraduate-level textbook that does otherwise ?

 

[Dale]

They are correct for an infinite plane with respect to the vertical component, which, in the example I give, would be σ/2 epsilon 0 with a ridiculously small error.

No, they aren't correct with respect to the vertical component.

To see this, consider the spot exactly in the center of a charged parallel plate capacitor (no external field other than the capacitor). In your calculation the magnitude of the vertical component of the E-field a small distance $\epsilon$ from the plate is the same above and below the plate. However, in the actual capacitor it is approximately zero on one side (the outside) and non-zero on the other side (the inside).

To get the approximately correct fields, you have to choose the correct boundary conditions. In this case the correct boundary condition is to add a uniform E-field in the vertical direction. This leads to a vertical component which does not behave like your solution, but is zero on one side and twice what you calculate on the other.

BOUNDARY CONDITIONS!!!

This is all completely standard stuff about boundary conditions. Well known, completely understood. It is simply usually ignored in the standard derivation. This is not a big breakdown of the model or a big problem of convergence or any of the other issues you are complaining about. You simply correctly have found that there are more solutions than the standard one, although you also have not found all of the solutions. You have to specify the boundary conditions to determine which solution is appropriate.

 

[A.T.]

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.

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

For the scenario above? What is that estimation based on?

 

[Vincf]

The parallel-plate capacitor is almost never treated as a boundary value problem. Aside from the rather limited conformal transformation, I only know of one article in the AJP that addresses it as such, and it's very complicated.
https://pubs.aip.org/aapt/ajp/artic...0/The-circular-disk-parallel-plate-capacitor?

Generally, it's treated as a summation problem. One define the charges and calculate the field.

For example, look at Griffiths, 4th edition,
p. 75:
Two infinite parallel planes carry equal but opposite uniform charge densities $\pm \sigma$.......
.....the left plate produces a field $(1/2 \epsilon_0) \sigma$.....:

No boundary value problem here. Just a simple summation.

and p. 106:
If we place $+Q$ on the top and $-Q$ on the bottom, they will spread out uniformly over the two surfaces provided the area is reasonably large and the separation small.

You can see that Griffiths positions himself far from the center without any problem (in his diagram, there is no center, just shaded planes). In the parallel plate capacitor, edge effects appear when we are at a distance for the edge is comparable to the distance between the edges. So, very far from the center!
This reasoning is very common, and I've always used the same one with my students (I'm not here to confuse them!), but I think it's at least incomplete.
The horizontal components of the field of each plate are not small, but since there are two opposite planes, they cancel each other out in pairs. The final conclusion is correct, even though the starting point is incomplete.

 

[Dale]

What do you think specifying a boundary value does?

 

[Vincf]

Solving the parallel-plate capacitor problem as a boundary value problem is actually very complicated. Take a look at the article I linked on the AJP website.
Even the conducting disk problem isn't easy to solve. If I remember correctly, the exercise is posed in the third edition of Volume II of Berkeley. And it's rather ingenious. I've also seen solutions using an unusual (elliptic?) coordinate system.
The advantage of formulating it as a boundary value problem is that the problem is very simple to state:
"solve Laplace's equation with the potential imposed on the conductors and being zero at infinity."
This works for all shapes of conductors, and if you have a numerical solver for Laplace's equation, it can be solved in a few seconds.
But for a physicist, the analytical solution has its charm!

 

[Dale]

What do you think specifying a boundary value does?

 

[PeroK]

That sounds very interesting. Do you have a reference handy for that?

The idea is that you have a right triangle with two unit sides. Shape n has n equal horizontal steps and n equal vertical steps instead of the hypotenuse. The length is always 2. But, the shapes tend to the triangle - by most measures of convergence (the steps converge uniformly to the hypotenuse as functions, for example).

All that means is that the perimeter of each shape does not tend to the length of the perimeter of the limit shape.

 

[PeroK]

This is why in the development of the Riemann integral, you need an sequence of upper bounds and a sequence of lower bounds that converge to the same limit. You can't rely on one-sided convergence, as it were.

 

[Dale]

What do you think specifying a boundary value does?

To move things along, I will answer and proceed.

Specifying a boundary value selects one solution out of a family of solutions.

Why it matters here: we are solving Maxwell’s equations, which are a set of partial differential equations. We can often use a lot of shortcuts and tricks to simplify things, but we are always going to produce a family of solutions that will require the specification of some boundary conditions to get a specific solution. You cannot avoid it. It is inherent in the fact that we are solving Maxwell’s equations.

Your solutions are valid solutions. The usual solution is a valid solution. They are all part of the same family of solutions to the same partial differential equations.

The only difference between your solution and the standard is the choice of boundary condition.

 

[haruspex]

It isn’t a convergence problem. None of the integrals or limits involved diverge.

Really?
In post #1, @Vincf obtained:

$$E_y(r_1, r_2) =- \frac{\sigma}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)$$
where $r_1, r_2$ are the distances from the reference point to the edges of the finite width infinite strip. If we treat the infinite plane as the sum of two semi-infinite planes we have $E_y=E_y(\infty, z)+E_y(z,\infty) =\infty-\infty$.
The divergence can be avoided by taking the $r_1, r_2$ limits simultaneously at a fixed ratio, but since the result depends on the ratio chosen it only helps prove the result is invalid.

 

[haruspex]

The idea is that you have a right triangle with two unit sides. Shape n has n equal horizontal steps and n equal vertical steps instead of the hypotenuse. The length is always 2. But, the shapes tend to the triangle - by most measures of convergence (the steps converge uniformly to the hypotenuse as functions, for example).

All that means is that the perimeter of each shape does not tend to the length of the perimeter of the limit shape.

A fractal example being the length of a coastline. Depends how long the ruler is.

 

[haruspex]

consider the spot exactly in the center of a charged parallel plate capacitor (no external field other than the capacitor). In your calculation the magnitude of the vertical component of the E-field a small distance ϵ from the plate is the same above and below the plate. However, in the actual capacitor it is approximately zero on one side (the outside) and non-zero on the other side (the inside).

Sorry, I am not following your argument.
@Vincf's calculation (in post #1.. maybe you are referring to a later post) is for a single charged sheet. A parallel plate capacitor has two oppositely charged sheets. The post #1 calculation extended to a capacitor would yield largely cancelling fields outside and reinforcing fields inside. What am I misinterpreting?

 

[Dale]

What am I misinterpreting?

You are missing the recent context. @Vincf is also interested in applying the infinite sheet of charge model to some real scenario. The capacitor discussion is in that context.

 

[haruspex]

You are missing the recent context. @Vincf is also interested in applying the infinite sheet of charge model to some real scenario. The capacitor discussion is in that context.

Ok, thanks.
So do you agree that the original problem is a convergence issue?