$A$ parallel plate capacitor is charged fully by using a battery. Then,without disconnecting the battery,the plates are moved further apart. Then,

$A$ 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 $,q$)

For the given circuit,find the voltage across the $2\ \mu F$ capacitor. (in $V$)

$A$ thin metallic partition of negligible thickness is inserted between two shaded metallic plates as shown. The remaining ends are then packed with insulating plates to form a container-like structure. Two taps shown are opened at $t = 0$ and finally closed at $t = 5 \ s$. Find the capacitance of the system between $A$ and $B$ after closing the taps. (Assume the liquid to be non-conducting). Volumetric flow rates and dielectric constants of the liquid are given.

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)