Source: High school physics (Chinese)
Problem
A proton with charge $+e$ and mass $m_p$ enters a uniform electric field between parallel plates with an initial velocity $v_0$. The velocity is perpendicular to the electric field lines. The plates have length $l$, are separated by a distance $d$, and have a voltage $U$ across them.
- Find the lateral displacement $y$ of the proton as it exits the field.
- An alpha particle ($q_\alpha = 2e, m_\alpha \approx 4m_p$) enters the same field. Find its lateral displacement $y_\alpha$ in terms of the proton's displacement $y_p$ if (a) it has the same initial velocity $v_0$, or (b) it has the same initial kinetic energy.
[Q1] $y = \frac{eUl^2}{2m_p d v_0^2}$ [Q2] (a) $y_\alpha = 0.5 y_p$, (b) $y_\alpha = 2 y_p$
The motion is projectile-like. The time spent in the field is $t = l/v_0$. The electric field is $E = U/d$. The lateral acceleration is $a_y = qE/m = qU/(md)$. The lateral displacement is $y = \frac{1}{2}a_y t^2$.
[Q1] For the proton ($q=e, m=m_p$):
$$y_p = \frac{1}{2} \left( \frac{eU}{m_p d} \right) \left( \frac{l}{v_0} \right)^2 = \frac{eUl^2}{2m_p d v_0^2}$$[Q2] The general expression for displacement is $y = \frac{qUl^2}{2mdv_0^2}$. For an alpha particle, $q_\alpha = 2e$ and $m_\alpha = 4m_p$.
(a) Same initial velocity $v_0$:
$$y_\alpha = \frac{(2e)Ul^2}{2(4m_p)dv_0^2} = \frac{1}{2} \frac{eUl^2}{2m_p d v_0^2} = \frac{1}{2}y_p$$(b) Same initial kinetic energy $K_0 = \frac{1}{2}mv_0^2$: The displacement can be written as $y = \frac{qUl^2}{4dK_0}$.
$$y_\alpha = \frac{q_\alpha Ul^2}{4dK_0} = \frac{(2e)Ul^2}{4dK_0} = 2 \left( \frac{eUl^2}{4dK_0} \right) = 2y_p$$