For the circuit shown in the figure,the charge on the $4 \mu F$ capacitor is $....... \mu C$.

$A$ 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 $V$?

$A$ charged capacitor is disconnected from the battery and if the distance between the two plates of the capacitor is increased then . . . . . . .

If the charge on a capacitor is increased by $2 \ C$,the energy stored in it increases by $21\%$. The original charge on the capacitor is....$C$

Two capacitors, $C_1$ and $C_2$ have their capacitances in the ratio $1:2$. $V_s$ and $V_p$ are the potential differences applied across the series and parallel combination of $C_1$ and $C_2$ respectively, so that the energy stored in the two cases becomes same. The ratio $V_s$ to $V_p$ is

$A$ parallel plate capacitor $A$ is filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to a voltage source $V_0 = 30\,V$ and then connected to the first capacitor $A$ after disconnecting the voltage source. Find the charge and voltage on capacitor $A$.