Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now, this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

$A$ capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of $6.0 \text{ V}$ and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$. (Assume capacitor $Q$ does not have any charge initially). The value of $\alpha$ is . . . . . . .

$A$ parallel plate capacitor of capacitance '$C$' is connected to a battery and charged to a potential '$V$'. Another capacitor of capacitance '$3C$' is charged to a potential '$3V$'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (in $CV^2$)

The energy of a charged capacitor is $U$. It is removed from the battery and then connected in parallel to another uncharged capacitor having capacitance twice that of the first one. The energy of the first and second capacitors respectively is . . . . . . .

$A$ capacitor of capacitance $0.2 \, \mu F$ is charged to a potential of $600 \, V$. After removing the battery,it is connected in parallel to an uncharged capacitor of $1.0 \, \mu F$. The final potential across the capacitors will be.........$V$.

Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.