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(B) $1$. Initial state: The capacitors $C$ and $2C$ are in series with battery $E$. The equivalent capacitance is $C_{eq} = \frac{C 	imes 2C}{C + 2C}

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$0, \frac{CE^2}{3}$

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(B) $1$. Initial state: The capacitors $C$ and $2C$ are in series with battery $E$. The equivalent capacitance is $C_{eq} = \frac{C 	imes 2C}{C + 2C} = \frac{2C}{3}$.$2$. The charge on each capacitor is $q = C_{eq}E = \frac{2CE}{3}$.$3$. Initial energy stored: $U_i = \frac{1}{2} C_{eq} E^2 = \frac{1}{2} (\frac{2C}{3}) E^2 = \frac{CE^2}{3}$.$4$. Final state: The capacitors are disconnected and reconnected with opposite polarity. The net charge on the plates connected together is $q - q = 0$. Since the total charge is zero,the final charge on each capacitor is $0$.$5$. Final energy stored: $U_f = 0$.$6$. Energy lost: $\Delta U = U_i - U_f = \frac{CE^2}{3} - 0 = \frac{CE^2}{3}$.$7$. Thus,the charge on $2C$ is $0$ and the energy lost is $\frac{CE^2}{3}$.

The given circuit shows two capacitors connected to a battery. After the capacitors are completely charged,the battery is removed and the capacitors are connected with plates of opposite polarity together. Then the charge on $2C$ and the energy lost in the process are respectively

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