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

$A$ $2 \mu F$ 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 . . . . . . $V$.

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

$A$ sphere of capacitance $100 \text{ pF}$ is charged to a potential of $100 \text{ V}$. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha \times 10^{-7} \text{ J}$. The value of $\alpha$ is . . . . . . . (combined capacitance of spheres is $200 \text{ pF}$)

Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12\, \mu C$ and $-3\, \mu C$ charges,respectively,and are at a large distance from each other. They are now connected by a conducting wire. $A$ long time after this is done,the charges on $S_{1}$ and $S_{2}$ are respectively: