Let the point at which the charged particle enters the electric field, be origin O (0, 0), then after travelling a horizontal displacement L, it gets deflected by displacement y in vertical direction as it comes out of electric field. So, co-ordinates of its initial position are $x_1$ = 0 and $y_1$ = 0 and final position on coming out of electric field are $x_2$ = L and $y_2$ = y Components of its acceleration are $a_x$ = 0 and $q_y$ = $\frac{F}{m}$ = $\frac{qE}{m}$ and of initial velocity are $u_x$ = $v_x$ and $u_y$ = 0 so, by 2nd equation of motion in horizontal direction, $y_2-y_1$ = $u_x t + \frac{1}{2} a_x t^{2}$ or L - 0 = $u_x t$ + 0 t = $\frac{L}{u_x}$ and by 2nd equation of motion in vertical direction, $y_2-y_1$ = $u_y t$ + $\frac{1}{2} a_y t^{2}$ or y - 0 = 0 + $\frac{1}{2} \cdot \frac{qE}{m} \cdot \left(\frac{L}{v_x}\right)^2$ ... (i) or y = $\frac{qEL^2}{2mv_x^2}$ This gives the vertical deflection of the particle at the far edge of the plate.