Charge on the capacitor in the given circuit in steady state condition is :- ............... $\mu C$

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $3C$ is similarly charged to a potential difference $3V$. The charging battery is then disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

$A$ particle of charge $Q$ and mass $M$ moves in a circular path of radius $R$ in a uniform magnetic field of magnitude $B$. The same particle now moves with the same speed in a circular path of same radius $R$ in the space between the cylindrical electrodes of a cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $3R/2$. Then the potential difference between the capacitor electrodes must be

$64$ drops,each having the capacity $C$ and potential $V$,are combined to form a big drop. If the charge on each small drop is $q$,then the charge on the big drop will be: (in $,q$)

In the given circuit,if the equivalent capacitance between $A$ and $B$ is $1\,\mu F$,what is the value of $C$?

$A$ capacitor $C_1 = 1\,\mu F$ can withstand a maximum voltage of $V_1 = 6\,kV$,and a capacitor $C_2 = 3\,\mu F$ can withstand a maximum voltage of $V_2 = 4\,kV$. If these two capacitors are connected in series,what is the maximum voltage (in $kV$) that can be applied to the combination?