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? I see no problem with claiming that zero is a solution, for the two-way infinite integral, for any function where $f(-x) = -f(x)$.

That works for the Cauchy Principal Value, but for neither the Riemann integral nor the Lebesgue integral.

 

[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.

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