Recent content by PeroK

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...

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: $$\int_0^{\infty} \frac 1 {1 + x^2} dx = \lim_{a\to 0}\bigg (\sum_{n = 1}^{\infty} \frac a {1 + (na)^2} \bigg )$$

I've started a new thread on how to do this properly: https://www.physicsforums.com/threads/electric-field-of-uniformly-charged-infinite-plane.1084230/

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}...

To many of us Elon Musk (and his kind) are the catastrophe that will destroy life for the vast majority of humans. There would be no escape on Mars. As Orwell put it: imagine a jackboot stamping on a human face - forever.

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...

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...

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...

Time is what a clock measures. In this context, at least.

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 =...

For example, the Newtonian gravitational force is given by ##F = \frac{Gm_1m_2}{r^2}##. In theory, two point particles can be arbitrarily close and the gravitational force arbitrarily large. But, the mathematics and the physical theory do not allow the limit of ##r = 0##.

The limit is not part of the theory of special relativity.

A clock can't travel at the speed of light. All motion between you and a clock is relative. If the clock is moving at 99% of the speed of light relative to you, then you are moving at 99% of the speed of light relative to the clock. Time dilation is relative and reciprocal. You measure the...

Check the rules. We can't discuss your personal speculations here.

Are you a flat Earther?