Interaction between a point charge and a ring

It is a fairly interesting problem, so I will try to be as clear as I can.

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$\bullet$ At a general position $\displaystyle x$ (very small),
Consider a Gaussian surface, which is a cylinder of height $\displaystyle 2\ell$ and radius $\displaystyle x$ .
$\bullet$ Let $E$ represent the Electric Field at the curved surface of the cylinder ( $\displaystyle E$ ) is constant as $\displaystyle x$ is small)

$\bullet$ Let $E'$ represent the Electric Field at the two flat surfaces of the cylinder ( $\displaystyle E'$ ) is constant as $\displaystyle x$ is small)

Considering the total Flux through the two flat surfaces of the cylinder,
$\displaystyle\phi_1 = 2\times (E'\times\pi x^2)$

As we know,
$\displaystyle E' = \frac{kQ\ell}{(R^2+\ell^2)^{\frac{3}{2}}} \approx \frac{kQ\ell}{R^3}$

Hence,

$\displaystyle\phi_1 = \frac{2\pi kQx^2}{R^3} \ell$

Considering the total flux through the curved surface of the cylinder,
$\displaystyle\phi_2 = E\times 2\pi x(2l)$

$\displaystyle\phi_2 = E\times(4\pi xl)$

Using Gauss's Law ,

$\displaystyle \phi_1+\phi_2 = \frac{q_{inside}}{\epsilon_o} = 0$ (Since there is no charge inside the Gaussian surface)

Hence, $\displaystyle\frac{2\pi kQx^2}{R^3} \ell = -E\times(4\pi xl)$

$\displaystyle\Rightarrow |E| = \frac{kQ}{2R^3} x$

$\displaystyle F = qE = \frac{kQq}{2R^3} x = Kx$ (say)

Thus,
$\displaystyle K = \frac{kQq}{2R^3}$

Using the concept of Reduced Mass ,
Reduced Mass $\displaystyle = \mu = \frac{Mm}{M+m}$

It is clear that,
$\displaystyle a = \frac{K}{\mu} x = w^2 x$

Thus,
$\displaystyle w = \sqrt{\frac{K}{\mu}} = \sqrt{\frac{2MmR^3}{KQq(M+m)}}$

$\Rightarrow\displaystyle T = \frac{2\pi}{w} = 2\pi\sqrt{\frac{2MmR^3}{KQq(M+m)}}$

The problem clearly asks us to find out $\displaystyle\frac{T}{2}$ , thus,
$\displaystyle T' = \frac{T}{2} = \pi\sqrt{\frac{2MmR^3}{KQq(M+m)}} \approx\boxed{0.0296sec}$

In the reference frame fixed to the system's centre of mass O, with x is the distance between point charge and centre of the ring, the distance from O to point charge: d {1}=x\frac{m}{M+m} centre of the ring: d {2}= x\frac{M}{M+m} Thus kinetic energy of each object is given by: K {1}= \frac{1}{2}M(d' {1})^ =\frac{1}{2}\frac{Mm^2}{(M+m)^2}x'^2 K {2} = \frac{1}{2}m(d' {1})^ =\frac{1}{2}\frac{mM^2}{(M+m)^2}x'^2 Interaction energy of system: (p is the linear charge density) W=\int\limits 0^{2\pi}{\frac{kqpRd\theta}{\sqrt{R^2+x^2-2Rxcos\theta}} =\int\limits 0^{2\pi}kqp (1+cos\theta\frac{x}{R}+ \frac{1+3cos^2\theta}{2}\frac{x^2}{R^2})d\theta = kqp(2\pi + \frac{\pi}{2}\frac{x^2}{R^2}) Law of energy conservation: W+K {1}+K {2} =const take derivative: \frac{kqp\pi}{R^2}x+\frac{Mm}{M+m}x''=0 the required time is half a revolution \tau=\pi\sqrt{\frac{mMR^2}{kqp(M+m)\pi}}=\pi\sqrt{\frac{2MmR^3}{kqQ(M+m)}}=0.0296s