Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?

Two identical capacitors each of capacitance $5 \mu F$ are charged to potentials $2 kV$ and $1 kV$ respectively. Their negative ends are connected together. When the positive ends are also connected together,the loss of energy of the system 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 . . . . . . .

$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another capacitor of the same capacity is connected in parallel to the first capacitor. Then the energy stored in each capacitor is

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential difference $V$ by connecting it to a battery. Another capacitor of capacitance $2C$ is charged to a potential difference $2V$ by connecting it to another battery. Now,the charging batteries are removed and the capacitors are connected in parallel such that the positive plate of one is connected to the negative plate of the other and the negative plate of the first is connected to the positive plate of the second. Find the final energy of this configuration.

$A$ $10\ \mu F$ capacitor is charged to a potential difference of $1000\ V$. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged $6\ \mu F$ capacitor. What is the final potential difference across each capacitor in $V$?