Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ is:

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

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

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?

$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.