Magnetic Or Oscillation?

Let the direction toward right side is positive side, then the magnetic force applied on electron at $B$ is

$F=kQ^2(\frac{1}{x^2}-\frac{1}{(3-x)^2})$

$ma=kQ^2(\frac{1}{x^2}-\frac{1}{(3-x)^2})$

$a=\frac{kQ^2}{m}(\frac{1}{x^2}-\frac{1}{(3-x)^2})$

When $x=1$ , $v=0$ , so when it travel to $x=1.5$ ,

$\int vdv=\int adx$

$\frac{v^2}{2}=\int_1^{\frac{3}{2}} \frac{kQ^2}{m}(\frac{1}{x^2}-\frac{1}{(3-x)^2}) dx$

$\frac{v^2}{2}=\frac{kQ^2}{6m}=\frac{Q^2}{24m\pi \varepsilon_0}$

$v^2=\frac{Q^2}{12m\pi \varepsilon_0}$

$v=\boxed{\frac{Q}{2\sqrt{3m\pi \varepsilon_0}}}$

Feel free to tell me if I had any wrong :)

Consider the total energy of the electron. It must be conserved. Initially, it has electric potential energy $\frac{Q^2}{4\pi\epsilon_0}\left(\frac{1}{1}-\frac{1}{2}\right)$ Initially, it has no kinetic energy as it is at rest. Now consider the total energy of the electron when it is at the centre. The electric potential energy is $\frac{Q^2}{4\pi\epsilon_0}\left(\frac{1}{1.5}+\frac{1}{1.5}\right)$ while it has kinetic energy $\frac{1}{2}mv^2$ . Equating, we get $v=\frac{Q}{2\sqrt{3m\pi\epsilon_0}}$ , giving us the desired answer of 5.