【題目】如圖所示，E＝10 V，r＝1 Ω，R1＝R3＝5 Ω，R2＝4 Ω，C＝100 μF。當(dāng)S斷開時(shí)，電容器中帶電粒子恰好處于靜止?fàn)顟B(tài)。求：

（1）S閉合后，帶電粒子加速度的大小和方向；

（2）S閉合后流過R3的總電荷量。

【答案】（1）g，方向豎直向上 （2）4×10－4 C

【解析】（1）開始帶電粒子恰好處于靜止?fàn)顟B(tài)，必有qE＝mg且qE豎直向上。S閉合后，qE＝mg的平衡關(guān)系被打破。

S斷開時(shí)，帶電粒子恰好處于靜止?fàn)顟B(tài)，設(shè)電容器兩極板間距離為d，有UC＝E＝4 V， ＝mg

S閉合后，UC′＝E＝8 V

設(shè)帶電粒子加速度為a，

則－mg＝ma，解得a＝g，方向豎直向上。

（2）S閉合后，流過R3的總電荷量等于電容器上電荷的增加量，所以ΔQ＝C（UC′－UC）＝4×10－4 C

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