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    【题目】如图所示，E＝10 V，r＝1 Ω，R1＝R3＝5 Ω，R2＝4 Ω，C＝100 μF。当S断开时，电容器中带电粒子恰好处于静止状态。求：

    （1）S闭合后，带电粒子加速度的大小和方向；

    （2）S闭合后流过R3的总电荷量。

    【答案】（1）g，方向竖直向上 （2）4×10－4 C

    【解析】（1）开始带电粒子恰好处于静止状态，必有qE＝mg且qE竖直向上。S闭合后，qE＝mg的平衡关系被打破。

    S断开时，带电粒子恰好处于静止状态，设电容器两极板间距离为d，有UC＝E＝4 V， ＝mg

    S闭合后，UC′＝E＝8 V

    设带电粒子加速度为a，

    则－mg＝ma，解得a＝g，方向竖直向上。

    （2）S闭合后，流过R3的总电荷量等于电容器上电荷的增加量，所以ΔQ＝C（UC′－UC）＝4×10－4 C

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