When two identical capacitors are charged individually to different potentials and connected in parallel to each other,after disconnecting them from the source:

$A$ 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 connected to another battery and is charged to a potential difference $2V$. The charging batteries are 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

$A$ capacitor has a capacitance of $15\ \mu F$ when filled with a dielectric of constant $K = 15$. Another capacitor has a capacitance of $1\ \mu F$ when filled with air. Both capacitors are charged by a $100\ V$ battery. After removing the battery and the dielectric from the first capacitor,the two capacitors are connected in parallel. What will be the final voltage across them in $V$?

Effective capacitance of parallel combination of two capacitors $C_{1}$ and $C_{2}$ is $10\; \mu F$. When these capacitors are individually connected to a voltage source of $1\; V,$ the energy stored in the capacitor $C_{2}$ is $4$ times that of $C_{1}$. If these capacitors are connected in series,their effective capacitance will be (in $; \mu F$)

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

$A$ parallel plate capacitor has capacitance $C$. If the charges on the plates are $Q$ and $-3Q$,find the potential difference between the plates.