Solution :
We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :
[tex]E_P=\int dE \cos[/tex]
[tex]E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}[/tex]
[tex]\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ[/tex]
[tex]\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}[/tex]
If we put an electron on point P, then force on point e is :
[tex]\vec{F}=-|e|\vec{E_P}[/tex]
[tex]F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}[/tex]
If r >> x , then [tex]$\frac{x^2}{r^2} \approx 0$[/tex]
Then, [tex]$\frac{-eKQ}{r^3}x$[/tex]
[tex]$ma =\frac{-eKQ}{r^3}x$[/tex]
[tex]$a =\frac{-eKQ}{mr^3}x$[/tex]
Compare, a = -ω²x
We get,
[tex]$\omega^2 = \frac{eKQ}{R^3m}$[/tex]
[tex]$\omega = \sqrt{\frac{eKQ}{r^3m}}$[/tex]
[tex]$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$[/tex]
[tex]$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$[/tex]