The vertical deflection $y$ of a charged particle moving perpendicularly into a uniform electric field is given by $y = \frac{1}{2} a t^2$. Since $a = \frac{qE}{m}$ and $t = \frac{x}{v_x}$, the deflection is $y = \frac{qE x^2}{2m v_x^2}$. This shows that the deflection $y$ is directly proportional to the charge-to-mass ratio ($\frac{q}{m}$), assuming the initial velocity and horizontal distance are comparable. From the figure (NCERT Physics Class 12, Fig 1.30), particle 3 suffers the maximum deflection, indicating it has the highest magnitude of charge to mass ratio. (See NCERT Physics Class 12, Exercise 1.13).