A uniform electric field E = ((8m)/(e))V m ^ - 1 is created between...

Time taken by the electron to cross will be, $t = \dfrac{1}{{v}_{x}} = \dfrac{1}{2}s$.

Acceleration of the electron along $y$ axis

${a}_{g} = \dfrac{F}{m} = \dfrac{Ee}{m} = \dfrac{8m}{e} \times \dfrac{e}{m} = 8m{s}^{- 2}$

${v}_{y}$ of the electron when it has crossed the capacitor will be,

${v}_{y} = {a}_{y}t = 8 \times \dfrac{1}{2} = 4m{s}^{- 1}$

Therefore,

$tan\theta = \dfrac{{v}_{y}}{{v}_{x}} = \dfrac{4}{2} = 2 ⇒ \theta = \tan^{- 1}\left(2\right)$