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I found the following question in a good physics book I was solving and although this is a computer problem, I wonder if it can be done without using computers.

$Q.$ Find the force of attraction between $2$ coaxial rings of radius $R$ having charge $+Q$ and $-Q$. Assume that the common axis is the $X-axis$ and the distance between their centres is $2R$.

First I assumed that the centre of the positivelt charged ring is $(0,0,0)$. Then I took a small element of length $Rd\theta$ on the positively charged ring which has co-ordinates $\vec\theta=(0,R\cos\theta,R\sin\theta)$ and another element on the negatively charged ring with length $Rd\phi$ and co-ordinates $\vec\phi=(2R,R\cos\phi,R\sin\phi)$. Hence the distance between the two elements is $$d=|\vec\phi-\vec\theta|=\sqrt{4R^2+2R^2-2R^2\cos(\theta-\phi)}=2R\sqrt{1+\sin^2\left(\frac{\theta-\phi}2\right)}$$ Now, by symmetry we can assume that the mutual force would be directed toward the centre and parallel to the common axis. Therefore first we have to find the component of the force between the 2 elements towards the centre. The angle between the line joining $\vec\theta$ and $\vec\phi$ and the line joining $\vec\phi$ and $\vec 0$ is given by $$\begin{align}\cos\alpha&=\frac{(\vec\phi-\vec\theta).\vec\phi}{|\vec\phi-\vec\theta||\vec\phi|}\\&=\frac{2+\sin^2\left(\frac{\phi-\theta}2\right)}{\sqrt5.\sqrt{1+\sin^2\left(\frac{\phi-\theta}2\right)}}\end{align}$$Hence the force on that element along the axis is$$F=\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{(Q/2\pi R)^2(2+\sin^2(\frac{\phi-\theta}2))}{4\pi \epsilon_{0}*4(1+\sin^2(\frac{\phi-\theta}2))*\sqrt5}*\frac2{\sqrt5} d\theta d\phi$$ The $\frac2{\sqrt5}$ comes because we are taking the component along the axis. The $R^2$ cancels in the numerator and denominator because of $Rd\theta$ and $Rd\phi$ and $R^2$ at the denominator. I am at a loss about how to evaluate this integral (It can be understood why it is a computer problem). So I have the following 3 doubts:-

$1.$ Is my approach correct?

$2.$ How can the integral be evaluated?

$3.$ Is there an easier way to calculate the required force?

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closed as off-topic by ACuriousMind, Kyle Kanos, AccidentalFourierTransform, user36790, CuriousOne Apr 7 '16 at 5:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Kyle Kanos, AccidentalFourierTransform, Community, CuriousOne
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ For #2, you might want to consider Mathematics instead. Note that we aren't a homework-help service, so #1 (and likely #3 as well) are considered off-topic here. You can make this on-topic by changing the question to ask about the physics concept that is giving you trouble. $\endgroup$ – Kyle Kanos Apr 6 '16 at 10:20
  • $\begingroup$ @KyleKanos, I understand what you are trying to say. But then how would you suggest I ask the above question in this forum... This is the only forum that answers physics doubts to my knowledge (I am a high school student... and I believe you can see that from my question)... you may edit the question to your likings (no objections) but at least keep the material of the question similar.... also could you give me an answer to the above in comments??? $\endgroup$ – Abhishek Bakshi Apr 9 '16 at 14:05
  • $\begingroup$ I would like to edit this post to get it reopened but I think there's a mistake in it. I think $|\vec{\phi} - \vec{\theta}|$ should be $R\sqrt{6 - 2 \cos(\theta - \phi)}$. How did you get the expression with the $\sin^2$? $\endgroup$ – DanielSank Oct 29 '16 at 18:18