$A$ capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when the medium between the plates is air. The charge on the plates becomes $100\,\mu C$ when the space between the plates is filled with oil. The dielectric constant of the oil is:

Two identical condensers are joined as shown in the figure. When the switch $S$ is closed,the total energy of the system is $U_1$. If the switch is opened and both the condensers are filled with a dielectric of dielectric constant $K = 3$,then the energy of the system becomes $U_2$. The value of $\frac{U_1}{U_2}$ is

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

$A$ parallel plate capacitor with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?

An air capacitor has a capacitance of $1 \mu F$. Now the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is: [$d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the first and second dielectric respectively] (in $\mu F$)

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?