Application of EM easy

Acc to Maxwell's equation, Gauss's law = $\oint E\cdot ds=Q$ is a complicated quantity. It won't be constant over any simple surface. $m\omega^2=\frac {d^2 v}{dx^2}|_{(0,0)}$ $\frac{d^2 v_{x,y}}{dx^2}|_{0,0}=\frac{d^2 v_{x,0}}{dx^2}|_{x=0}$ $V_{x,0}=\int\frac{dq'}{r'}\cdot\frac {1}{4\pi\epsilon_0}=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{q\lambda rd\theta'}{(r^2+x^2+2rxCos\theta')}$ $=\frac {q\lambda}{4\pi\epsilon_0}\int_0^{2\pi}(1+2\frac {x}{r}Cos\theta'+(\frac{x}{r})^2)^{-\frac {1}{2}}d\theta'$ $\frac{x}{r}\approx$ small $V_{(x,0)}=\frac {q\lambda}{4\pi\epsilon_0}\int_0^{2\pi}d\theta [1-Cos\theta(\frac {x}{r})+\frac{3Cos^2\theta-1}{2}(\frac {x^2}{r^2})+\theta (\frac{x^2}{r^2})]$ $\frac{d^2 v_{x,0}}{dx^2}|_{x=0}=\frac {q\lambda}{4\pi\epsilon_0}\int_0^{2\pi}d\theta\frac {3Cos^2\theta-1}{2}\cdot\frac {2}{r^2}$ $=\frac{q\lambda}{4\epsilon_0 r^2}$ $\omega={\sqrt{\frac{\frac{d^2 v}{dx^2}|_{0,0}}{m}}}$ $\omega={\sqrt {\frac {q\lambda}{4m\epsilon_0 r^2}}}$