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.


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