Mutual E force due to charged coaxial rings
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. 521
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? 402