A uniform electric field of 10 N C$$^{-1}$$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle ($$\theta$$) of deviation of the path of electron as it comes out of the field is _____ (in degree).

$$L = v_x t \implies t = \frac{L}{v_x}$$

$$\frac{1}{2}m v_x^2 = 0.5e \implies v_x = \sqrt{\frac{e}{m}}$$

$$v_y = a_y t = \left( \frac{eE}{m} \right) t$$

$$v_y = \left( \frac{eE}{m} \right) \left( \frac{L}{v_x} \right)$$

$$\tan \theta = \frac{v_y}{v_x}$$

$$\frac{v_y}{L} = \frac{eE}{m v_x} \implies \tan \theta = \frac{eE}{m v_x} \cdot \frac{L}{v_x} = \frac{eE}{m} \cdot \frac{L}{v_x^2}$$

$$\tan \theta = \frac{eE}{m} \cdot \frac{L}{(e/m)} = E \cdot L$$

$$\tan \theta = 10 \times 0.1 = 1$$

$$\theta = \tan^{-1}(1) = 45^\circ$$