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 $\mu C$?

Consider the situation shown in the figure. Capacitor $A$ has a charge $q$ on it,whereas $B$ is uncharged. The charge appearing on capacitor $B$ a long time after the switch $S$ is closed is

Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

$A$ $0.2\,F$ capacitor is charged to $600\,V$ by a battery. On removing the battery,it is connected to another parallel plate capacitor of $1\,F$. The potential decreases to.........$V$.

$A$ 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:

$A$ $2\, \mu F$ capacitor $C_{1}$ is first charged to a potential difference of $10\, V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of $8\, \mu F$. The charge in $C_{2}$ on equilibrium condition is $\ldots\, \mu C$. (Round off to the Nearest Integer)