$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.

In the following figure,$C_1=5 \mu F$,$C_2=C_3=10 \mu F$ and $\varepsilon=20 \text{ V}$. Initially,the switch $S$ is connected to point $A$ until capacitor $C_1$ is fully charged. Afterwards,the switch is thrown to the left side and connected to point $B$. The charge on capacitor $C_3$ after equilibrium is reached will be (in $\mu C$)

Two capacitors of $100 \mu F$ and $50 \mu F$ are connected in parallel. If the potential difference across $100 \mu F$ is $20 \text{ V}$ and across $50 \mu F$ is $40 \text{ V}$,then the common potential of the parallel combination will be (assuming same polarities of the capacitors are connected together).

Two spheres of radii $1 \, cm$ and $2 \, cm$ are charged with $1.5 \times 10^{-8} \, C$ and $0.3 \times 10^{-7} \, C$ of positive charge,respectively. When they are connected by a wire,the charge will flow:

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 . . . . . . .

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is