根据题目和答案解析，生产讲解视频。1.视频需要有声音；2.需要根据解答步骤生成视频；3.视频总需要展示出电子的运动轨迹。---Here is the extraction of the content from the image:
**Problem Description:**
1. 如图所示是一个示波管工作原理图，电子(电量为e、质量为m)经电压U₁加速后，垂直进入间距为d、板长L的平行金属板组成的偏转电场，离开电场后打在距离偏转电场s的屏幕上的P点，已知偏转电场电压为U₂, 求:
**Sub-questions:**
(1)电子进入偏转电场时的速度v₀;
(2)打在屏幕上的侧位移y_OP;
**Diagram Description:**
* Type: Schematic diagram illustrating the working principle of an oscilloscope or cathode ray tube.
* Main Elements:
* **Accelerating Region:** Represented by vertical plates labeled U₁. Shows an electron beam entering from the left and being accelerated horizontally, exiting with initial velocity v₀.
* **Deflection Region:** Consists of a pair of horizontal parallel metal plates of length L. An electron path is shown bending downwards within these plates. The horizontal distance between the plates is L. The vertical distance between the plates is labeled implicitly as d in the solution context.
* **Drift Region:** The space between the deflection plates and the screen, with horizontal distance s.
* **Screen:** A vertical screen on the right.
* **Points:** Point O on the screen represents the undeflected position, directly in line with the entry point into the deflection plates. Point P on the screen is the actual point where the deflected electron hits. P is below O.
* **Paths:** An initial horizontal path leading to the deflection plates. A curved (parabolic) path within the deflection plates. A straight line path after leaving the deflection plates, hitting P. A horizontal dashed line extends from the entry point of the deflection plates through O on the screen.
* **Labels:** U₁ (accelerating voltage), v₀ (velocity entering deflection field), L (length of deflection plates), s (distance from plates to screen), O, P.
**Solution:**
[答案] 解:
(1)设电子经加速电场U₁加速后以速度v₀进入偏转电场，
由动能定理有: eU₁ = (1/2)mv₀²,
得: v₀ = sqrt(2eU₁/m)
(2)进入偏转电场粒子做类平抛运动，在电场方向有: a = eE/m = e(U₂/d)/m = eU₂/md
由水平分运动可得电子在电场中飞行时间: t₁ = L/v₀
电场方向的速度v_y = at₁ = (eU₂/md) * (L/v₀) = eU₂L/(mdv₀)
设射出偏转电场时速度的偏角为θ，则 tanθ = v_y/v₀ = (eU₂L/(mdv₀)) / v₀ = eU₂L/(mdv₀²)
速度的反向延长线恰好平分水平方向的位移，故 tanθ = OP / (s + L/2)
得 OP = U₂L(2s+L)/(4dU₁)。