In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ 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 finally on $C_2$ in $\mu C$?

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging batteries are then removed,and the capacitors are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration 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 hollow metallic spheres of radii $20 \ cm$ and $10 \ cm$ each carry a charge of $150 \ \mu C$. If they are connected by a conducting wire,the common potential is ...... $V$.

$A$ 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$)

Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?