$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

$A$ 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 $CV^2$)

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be

$A$ 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 capacitors $C_1$ and $C_2$ are connected in parallel. If a charge $Q$ is given to the combination,the charge gets shared. Then the ratio of charge on $C_1$ to charge on $C_2$ is . . . . . . .

Two conducting spheres of radii $5\, cm$ and $10\, cm$ are given a charge of $15\,\mu C$ each. After the two spheres are joined by a conducting wire, the charge on the smaller sphere is.......$\mu C$.