The charge on each capacitor when a voltage source of $15 \ V$ is connected in the circuit as shown,is (in $\mu C$)

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)

Find the equivalent capacitance across $AB$ in $\mu F$ (all capacitances are in $\mu F$).

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:

$A$ 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:

Three identical charged capacitors each of capacitance $5 \,\mu F$ are connected as shown in the figure. The potential difference across capacitor $(3)$,a long time after the switches $K_1$ and $K_2$ are closed,is ............ $V$.