A uniform electric field E is created between two parallel charged plates as shown in figure. An electron enter the field symmetrically between the plates with a speed u . The length of each plate is $l$. Find the angle of deviation, in degree of the path of the electron as it comes out of the field for $u^2=\frac{\sqrt{3}Ele}{m}$

Along the X – direction speed of electron remains uniform , because of no component of field along horizontal, 

So $x=ut$

$\mathcal{l}=ut$

Along y direction

$u_y=0$

Acceleration along y – direction of motion of electron is given by

$a_y=a=\frac{eE}{m}$

Since $v_{\gamma }=u_{\gamma }+a_{\gamma }t$

$\Rightarrow v_{\gamma }=0+\left(\frac{eE}{m}\right)t$

$From\left(1\right),wegett=\frac{\mathcal{l}}{u}$

$V_{\gamma }=\left(\frac{eE}{m}\right)\left(\frac{\mathcal{l}}{u}\right)$

$If\theta$ be the angle of deviation, them

$\tan \theta =\frac{v_{\gamma }}{v_{\gamma }}=\frac{\left(\frac{eE\mathcal{l}}{mu}\right)}{u}$

$\theta ={\tan }^{-1}\left(\frac{eE\mathcal{l}}{m{u}^2}\right)$

$\Rightarrow \theta ={\tan }^1\left[\frac{eE\mathcal{l}}{m\left(\sqrt{3}\frac{eE\mathcal{l}}{m}\right)}\right]={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^0$