An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

Two metallic plates form a parallel plate capacitor. The distance between the plates is $d$. $A$ metal sheet of thickness $\frac{d}{2}$ and of area equal to the area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

$A$ thin metal plate $P$ is inserted halfway between the plates of a parallel plate capacitor of capacitance $C$ in such a way that it is parallel to the two plates. The capacitance now becomes

$A$ parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $k(x)=K(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d) << 1$,the total capacitance of the system is best given by the expression:

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ are filled between the plates of a parallel plate capacitor as shown in the figure. If the area of each plate is $A$ and the separation between the plates is $d$,what is the equivalent capacitance of the capacitor?

$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?