Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

$A$ $2 \mu F$ capacitor is connected to a $50 \ V$ supply,and a $3 \mu F$ capacitor is connected to a $100 \ V$ supply. After removing the batteries,if the two plates of the same type of charge are connected to each other,the new potential difference is . . . . . . $V$.

$A$ 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 . . . . . . .

The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected by a thin wire,then the loss of energy will be given by