Source: High school physics (Chinese)
Problem
Two charged particles enter a charged parallel-plate capacitor with velocities parallel to the plates. Gravity is neglected. Both particles exit through the far end of the capacitor with the same deflection angle. Which of the following initial conditions must they satisfy?
(5) only.
Inside the plates, the time of flight is $t = L/v_0$ (with $L$ the plate length and $v_0$ the initial horizontal velocity). The vertical velocity gained is $v_y = (qE/m) t = qEL/(m v_0)$. The deflection angle satisfies
$$\tan\theta = \dfrac{v_y}{v_0} = \dfrac{qEL}{m v_0^2}.$$Hence equal deflection requires $\dfrac{q}{m v_0^2}$ to be the same for both particles.
(1) Same $q$ alone does not constrain $m v_0^2$. \ (2) Same $v_0$ alone does not constrain $q/m$. \ (3) Same $p = m v_0$ gives $m v_0^2 = p v_0$; with $p$ fixed but $v_0$ unconstrained, $q/(m v_0^2)$ varies. \ (4) Same $\tfrac{1}{2} m v_0^2$ constrains $m v_0^2$ but still leaves $q$ free. \ (5) If both particles are accelerated from rest through a common accelerating voltage $U_{\text{acc}}$, then $\tfrac{1}{2} m v_0^2 = q U_{\text{acc}}$, so $m v_0^2 = 2 q U_{\text{acc}}$, giving
$$\dfrac{q}{m v_0^2} = \dfrac{1}{2 U_{\text{acc}}},$$which is the same for both particles, regardless of $q$ or $m$. Hence (5) guarantees equal deflection.