In the circuit shown in the figure,$K$ is open. The charge on capacitor $C$ in the steady state is $q_1$. Now,the key $K$ is closed,and at the steady state,the charge on $C$ is $q_2$. The ratio of charges $\left( \frac{q_1}{q_2} \right)$ is:

Three capacitors are connected to a battery as shown in the figure. The ratio of charge on capacitors $C_3$ and $C_1$ is (in $.5$)

In the following figure,the charge on each capacitor in the steady state will be.....$\mu C$.

Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

What is the total electrostatic potential energy of the given system in $J$? (Given: $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} \ N \cdot m^2/C^2$)

$A$ capacitor $A$ has a capacitance of $15\ \mu F$ when filled with a dielectric of constant $K = 15$. Another capacitor $B$ is air-filled and has a capacitance of $1\ \mu F$. Both are charged to $100\ V$. After charging, the dielectric is removed from capacitor $A$, and the two capacitors are connected in parallel. The common potential difference across them will be ... $V$.