In the given circuit,if the equivalent capacitance between $A$ and $B$ is $1\,\mu F$,what is the value of $C$?

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance between points $x$ and $y$ is (in $\mu F$)

Calculate the charge on the capacitor in the steady state. (in $\mu C$)

$A$ composite parallel plate capacitor is made up of two different dielectric materials with different thicknesses ($t_{1} = 0.5 \text{ mm}$ and $t_{2} = 1 \text{ mm}$) and dielectric constants ($\epsilon_{r1} = 3$ and $\epsilon_{r2} = 4$) as shown in the figure. The two dielectric materials are separated by a conducting foil $F$. The voltage of the conducting foil is $.....V$.