电子从加速电场$U_{1}$飞出后的水平速度$v_{0}$大小？

$\\dfrac{2L}{T}$

粒子在加速电场中则$U_{1}e= \dfrac{1}{2}mv_{0}^{2}$

其中$U_{1}=\dfrac{2 mL^{2}}{eT^{2}}$

解得$v_{0}=\dfrac{2L}{T}$

$t=0$时刻射入电容器的电子离开电场时偏离$AB$中轴线的距离$y$；

$\\dfrac{U_{0}eT^{2}}{8md}$

粒子在偏转电场中水平方向匀速运动，则时间$t=\dfrac{2L}{v_{0}}=T$

加速度$a_{1}=\dfrac{U_{0}e}{md}$

则$0 \sim \dfrac{T}{2}$竖直位移$y_{1}=\dfrac{1}{2}a_{1}{\left(\dfrac{T}{2}\right)}^{2}=\dfrac{U_{0}eT^{2}}{8md}$

$\dfrac{T}{2} \sim T$时间内做减速运动，加速度$a_{2}=\dfrac{2U_{0}e}{md}=2a_{1}$

则$y_{2}=\left(a_{1}\dfrac{T}{2}\right)\dfrac{T}{2}-\dfrac{1}{2}a_{2}{\left(\dfrac{T}{2}\right)}^{2}=0$

电子离开电场时偏离$AB$中轴线的距离$y=y_{1}+y_{2}=\dfrac{U_{0}eT^{2}}{8md}$

在足够长的时间内从中线上方离开偏转电场的电子占电子总数的比值。

$\\dfrac{1}{6}$

设从$t_{1}$时刻进入电容器。$\dfrac{1}{2}a_{1}{\left(\dfrac{T}{2}-t_{1}\right)}^{2}+\left\{ a_{1}\left( \dfrac{T}{2}-t_{1} \right)\dfrac{T}{2}-\dfrac{1}{2}2a_{1}{\left(\dfrac{T}{2}\right)}^{2}\right\}+\left\{ \left(-\dfrac{T}{2}a_{1}-a_{1}t_{1} \right)t_{1}+\dfrac{1}{2}a_{1}t_{1}^{2} \right\}=0$

可得$t_{1}=\dfrac{T}{12}$

设从$T$时刻前的一段时间$t_{2}$进入电容器。$\dfrac{1}{2} \times 2a_{1}t_{2}^{2}+\left\{ 2a_{1}t_{2}\dfrac{T}{2}-\dfrac{1}{2}a_{1}{\left(\dfrac{T}{2}\right)}^{2}\right\}+\left\{\left( 2a_{1}t_{2}-a_{1}\dfrac{T}{2} \right)\left( \dfrac{T}{2}-t_{2} \right)+\dfrac{1}{2} \times 2a_{1}{\left(\dfrac{T}{2}-t_{2}\right)}^{2}\right\}=0$

$t_{2}=\dfrac{T}{12}$

综上，在足够长的时间内从中线上方离开偏转电场的电子占电子总数的比值为$\dfrac{\dfrac{T}{12}+\dfrac{T}{12}}{T}=\dfrac{1}{6}$