# Happy physics time

In this post, I consider a bar of infinite length with a finite charge, a charged particle, and examine what is the electromagnetic pull between the two. The situation is no more fuzzy than this:

Given a point $x$ on the $x$-axis (representing the charged bar), the distance between it and the charged particle $Q_2$ is given by

$\displaystyle d(x) = \sqrt{d^2 + x^2},$

where $d$ is the distance of the charged particle from the charged bar. By Coulomb’s law the scalar force between two charged objects with charges $Q_1$ and $Q_2$ is

$\displaystyle F = \frac{k_{\varepsilon}Q_1 Q_2}{r^2},$

where $k_{\varepsilon}$ is the Coulombs’s constant, and $r$ is the distance between the two particles.

We assume that the charged bar has constant charge density: all subchunks of equal length has equal charge. Now it is easy to see that if basic bar component of length $\Delta x$ has the charge of $\Delta x Q / w$, we can write

\displaystyle \begin{aligned} F &= \sum \Delta F \\ &= \sum \frac{k_{\varepsilon} Q_1 \frac{\Delta x}{w} Q_2}{d^2(x)} \\ &= \int_{-w/2}^{w/2} \frac{k_{\varepsilon} Q_1 Q_2 \mathrm{d}x}{wd^2(x)} \\ &= \frac{k_{\varepsilon}Q_1 Q_2}{w} \int_{-w/2}^{w/2} \frac{\mathrm{d}x}{x^2 + d^2} \\ &= \frac{k_{\varepsilon}Q_1 Q_2}{w} \Bigg[ \frac{\arctan(x / d)}{d} \Bigg]_{x = -w / 2}^{x = w / 2} \\ &= \frac{k_{\varepsilon}Q_1 Q_2}{dw} \Bigg[ \arctan(x / d) \Bigg]_{x = -w / 2}^{x = w / 2} \\ &= \frac{2k_{\varepsilon}Q_1 Q_2}{dw} \arctan(\frac{w}{2d}) \end{aligned}

Since $d, Q_1, Q_2$ and $k_{\varepsilon}$ are considered constants, the above expression approaches zero as $w \rightarrow \infty$.