Electric Field of Uniformly Charged Infinite Plane

[PeroK]

Here is a valid derivation of the electric field for an infinite plane, based on the result for an infinite wire of uniform linear charge density.
The electric field for an infinite wire of uniform charge density $\lambda$ at a distance $h$ from the wire is:
$$E_l = \frac 1 {2\pi \epsilon_0} \big (\frac \lambda h \big )$$Consider the plane as a infinite sequence of parallel uniformly charged wires, separated by a distance $d$ from each other. The component of the electric field in the direction away from the plate is given by:
$$E_n = \frac 1 {2\pi \epsilon_0} \big (\frac {\lambda h}{h^2 + n^2d^2} \big )$$Where the nth wire is a distance $nd$ from some reference point on the plane. And the field is measured a height $h$ above this point. The total electric field at this point is:
$$E = E_0 + 2\sum_{n = 1}^{\infty} E_n$$To approximate a uniformaly charge plane of charge density $\sigma$, we need to take $d \to 0$. And, for a give $d$, we need to set $\lambda = \sigma d$. This gives the required component of the electric field as:
$$E = \frac 1 {2\pi \epsilon_0}\big [ \frac {\sigma d}{h} + 2\sum_{n = 1}^{\infty} \frac{\sigma d h}{h^2 + n^2d^2} \big ]$$$$= \frac 1 {2\pi \epsilon_0}\big (\frac {\sigma d}{h} \big ) \big [1 + 2\sum_{n = 1}^{\infty} \frac{1}{1 + n^2(d/h)^2} \big ]$$Now we need to take the limit as $d \to 0$ (which is equivalent to taking $a = \frac d h \to 0$
$$E = \frac{\sigma}{\pi \epsilon_0} \lim_{a \to 0}\bigg (\sum_{n = 1}^{\infty} \frac{a}{1 + n^2a^2} \bigg )$$Finally, all we have to do is show that:
$$\lim_{a \to 0}\bigg (\sum_{n = 1}^{\infty} \frac{a}{1 + n^2a^2} \bigg ) = \frac \pi 2$$And we have, as required:
$$E = \frac {\sigma}{2\epsilon_0}$$

 

[Vincf]

Hello,
The sum you're showing is a Riemann sum. The Riemann sum is $$\sum_{n=0}^\infty f(x_{n})(x_{n+1}- x_{n}) $$ with the function $$f(x)=\frac {1} {1+x^2}$$ and $x_{n}=na$.

You do indeed arrive at the function $arctan(\infty)=\pi /2$
But you're only calculating the vertical component, which doesn't present any convergence problems.

For the horizontal component, you would ultimately find a Riemann sum corresponding to the integral of the calculation I proposed. If you sum symmetrically, you'll find 0. But if you take the limits separately, you should encounter the same convergence problems I've tried to demonstrate.

 

[PeroK]

Re the above identity for $\pi$.
$$\int_0^{\infty} \frac 1 {1 + x^2} dx = \bigg [\tan^{-1}(x)\bigg ]_0^{\infty} = \frac \pi 2$$And , from the definition of the Riemann integral:
$$\int_0^{\infty} \frac 1 {1 + x^2} dx = \lim_{a\to 0}\bigg (\sum_{n = 1}^{\infty} \frac a {1 + (na)^2} \bigg )$$

 

[PeroK]

For the horizontal component, you would ultimately find a Riemann sum corresponding to the integral of the calculation I proposed. If you sum symmetrically, you'll find 0. But if you take the limits separately, you should encounter the same convergence problems I've tried to demonstrate.

The horizontal component is zero, by symmetry. I've not got time to do the asymmetric calculation now. I'll do it later.

Although, I'm struggling to see how it can go wrong, whatever way things are done!

I would need to look carefully at your OP in the other thread, to see where things went awry.

 

[Dale]

I would need to look carefully at your OP in the other thread, to see where things went awry.

I don’t think they did go awry. It is a valid solution to Maxwell’s equations. It is just a different boundary condition than usual.

 

[Vincf]

For the usual infinite plane, we cannot impose the usual boundary conditions. The field does not tend towards 0 at infinity, and the potential diverges. This is, in fact, part of the problem: we extend the Coulomb integral to an infinite distribution without worrying about the convergence of the integrals.

 

[Dale]

For the usual infinite plane, we cannot impose the usual boundary conditions.

By “usual” I meant “usual for the infinite plane”.

Most people, when solving the infinite plane problem, implicitly choose a boundary condition such that the solution is perpendicular to the plane and equal magnitude on either side of the plane. That is a solution to Maxwell’s equations for the plane of charge.

But that is just one of a family of solutions. You can also add any constant uniform E field throughout space. That gives you another valid solution.

Your family of solutions is one where the constant uniform E field is aligned with the $x$ axis. It is not the most general family of solutions, but it is more general than the usual one.