A dejected charged particle!

Relevant wiki: Using Gauss' law to find E-field and capacitance

Electric field due to an infinitely charged wire at a distance $x$ from it measured perpendicular to it :

$\begin{aligned} \boxed{E=\dfrac{\lambda}{2\pi \epsilon_{0} x}} && \small \left( \text{ Refer to the wiki} \right) \end{aligned}$

Total force on the particle due to the charged wire :

$\begin{aligned} & F=\dfrac{q\lambda}{2\pi \epsilon_{0} x} \\&\Rightarrow ma=\dfrac{q\lambda}{2\pi \epsilon_{0} x} \\ & \Rightarrow m v.\dfrac{\mathrm{d}v}{\mathrm{d}x}=\dfrac{q\lambda}{2\pi \epsilon_{0} x} & \small \left(\color{#3D99F6}{a\text{ is the acceleration of the particle and } a=v.\dfrac{\mathrm{d}v}{\mathrm{d}x}} \right) \\& \Rightarrow mv.\mathrm{d}v=\dfrac{q\lambda.\mathrm{d}x}{2\pi \epsilon_{0} x} & \cdots (1) \\ &\Rightarrow m\displaystyle{ \int^{2u}_{u}{v}\mathrm{d}v}=\dfrac{q\lambda}{2\pi \epsilon_{0}}\displaystyle{\int^{2L}_{L}{\dfrac{\mathrm{d}x}{x}}} \\ &m \left[\dfrac{3u^{2}}{2} \right]=\dfrac{q\lambda}{2\pi \epsilon_{0}}\cdot \left[\ln{2}\right] & \cdots(2) \end{aligned}$

Using equation $(1)$ we can have :

$\begin{aligned} & m\displaystyle{ \int^{v_{f}}_{u}{v}\mathrm{d}v}=\dfrac{q\lambda}{2\pi \epsilon_{0}}\displaystyle{\int^{4L}_{L}{\dfrac{\mathrm{d}x}{x}}} \\ & \Rightarrow m \left[\dfrac{v_{f}^{2}-u^{2}}{2}\right] =\dfrac{q\lambda}{2\pi \epsilon_{0}}\cdot \left[\ln{4}\right] & \cdots(3) \end{aligned}$

Now using equation $(2)$ and $(3)$ we get the value of $v_{f}$ as follows :

Divide equation $(3)$ by equation $(2)$ to get :

$\dfrac{v_{f}^{2}}{3u^{2}}-\dfrac{1}{3}=2 \Rightarrow \boxed{ v_{f}=u\sqrt{7}}$