In the given circuit, the ratio of charge stored in the capacitors $C, 2C, 3C, 4C, 5C$ is

$A$ capacitor has a capacitance of $15\ \mu F$ when filled with a dielectric of constant $K = 15$. Another capacitor has a capacitance of $1\ \mu F$ when filled with air. Both capacitors are charged by a $100\ V$ battery. After removing the battery and the dielectric from the first capacitor,the two capacitors are connected in parallel. What will be the final voltage across them in $V$?

The following figure shows a $9 \ V$ battery and $3$ uncharged capacitors of capacitances $C_1 = C_2 = C_3 = 1 \ \mu F$. The switch is thrown to the right side until capacitor $C_1$ is fully charged,then the switch is thrown to the left. The final charge on capacitor $C_2$ is (in $\mu C$)

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

$A$ parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then

$27$ similar drops of mercury are maintained at $10 \, V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ....... times that of a smaller drop.