$64$ 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 $,q$)

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now 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

The distance between the plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs,one of a conductor and the other of a dielectric,both of equal thickness $d$,are inserted between the plates as shown in the figure. The potential versus distance graph will look like:

The charge on any one of the $2\,\mu F$ capacitors and the $1\,\mu F$ capacitor will be given respectively (in $\mu C$) as

$A$ flat circular disc has a charge $+Q$ uniformly distributed on it. $A$ charge $+q$ is thrown with kinetic energy $E$ towards the disc along its normal axis. The charge $q$ will:

Two conducting shells of radius $a$ and $b$ are connected by a conducting wire as shown in the figure. The larger shell is grounded. The capacitance of the system is: