有一道关于带电电容器的连接物理竞赛题,题目如下: 例1 在图1所示的电路中,直流稳压电源的电压U=6伏,电阻R=5欧,电容器C_1=1微法,C_2=2微法,若先使电键K闭合,然后再将K断开,则断开后C_1两极板间的电压为____伏,C_2两极板间的电压为____伏。 分析 K断开前C_1带电量q_(10)=C_1U=6微库,C_2不带电,K断开后,C_1与C_2串联,多数考生仍按原来不带电的电容器串联的分压规律,得到 There is a connection physics contest on charged capacitors. The topics are as follows: Example 1 In the circuit shown in Figure 1, the voltage of the DC stabilized power supply is U=6V, the resistance is R=5Ω, and the capacitor C_1=1μF, C_2. = 2 microfarads. If the key K is closed before K is disconnected, the voltage between the C_1 plates is ____ volts and the voltage between the C_2 plates is ____ volts. Analysis K disconnected before the C_1 charge q_ (10) = C_1U = 6 micro-library, C_2 is not charged, K disconnected, C_1 and C_2 in series, most candidates still according to the original uncharged capacitor series voltage divider law, obtained