$A$ parallel plate capacitor of capacitance $500 \ pF$ is charged with a $100 \ V$ supply. It is then disconnected from the supply and connected to another uncharged $500 \ pF$ capacitor. The electrostatic energy lost in this process is (in $\mu J$)

In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

Two spheres $A$ and $B$ of radii $4 \text{ cm}$ and $6 \text{ cm}$ are given charges of $80 \mu\text{C}$ and $40 \mu\text{C}$ respectively. If they are connected by a fine wire,the amount of charge flowing from one to the other is

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

One of the two identical capacitors having the same capacitance $C$ is charged to a potential $V_1$ and the other is charged to a potential $V_2$. If they are connected with their like plates together,then the decrease in the electrostatic potential energy of the combined system is

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