In the circuit shown in the figure,the ratio of charges on $5 \mu F$ and $4 \mu F$ capacitors 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 $V$?

In the given circuit,$C_1=2\,\mu F, C_2=0.2\,\mu F, C_3=2\,\mu F, C_4=4\,\mu F, C_5=2\,\mu F, C_6=2\,\mu F$. The charge stored on capacitor $C_4$ is $.....\mu C$.

The resultant capacitance between $A$ and $B$ in the figure is ...... $\mu F$.

Find the equivalent capacitance between $A$ and $B$.

$A$ source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key $K$ is closed,the total energy stored across the combination is $E_{1}$. Now key $K$ is opened and a dielectric of dielectric constant $K_{d} = 5$ is introduced between the plates of both capacitors. The total energy stored across the combination is now $E_{2}$. The ratio $E_{1} / E_{2}$ will be: