$A$ capacitor of capacitance $C_1 = 1 \ \mu F$ is charged using a $9 \ V$ battery. $C_1$ is then removed from the battery and connected to capacitors $C_2$ and $C_3$ of $2 \ \mu F$ and $3 \ \mu F$ respectively,as shown in the figure. Find the charge on $C_3$ after equilibrium is reached.

$A$ capacitor having capacitance $C$ is charged to a voltage $V$. It is then removed and connected in parallel with another identical capacitor which is uncharged. The new charge on each capacitor is now

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

$A$ 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 $CV^2$)

Energy of a charged capacitor is $U$. Now it is removed from a battery and then connected to two other identical uncharged capacitors in parallel. What will be the energy of each capacitor?

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch and a resistor 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: