When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......

$A$ parallel plate capacitor with air medium between the plates has a capacitance of $10 \mu F$. The area of the capacitor is divided into two equal halves and filled with two media (as shown in the figure) having dielectric constants $K_1=2$ and $K_2=4$. The capacitance of the system will be (in $\mu F$)

$A$ parallel plate capacitor has plate area $40 \text{ cm}^2$ and plate separation $2 \text{ mm}$. The space between the plates is filled with a dielectric medium of thickness $1 \text{ mm}$ and dielectric constant $K = 5$. The capacitance of the system is:

$A$ 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:

$A$ parallel plate capacitor having plates of area $S$ and plate separation $d$,has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $(\varepsilon_1=2$ and $\varepsilon_2=4)$ are introduced between the two plates as shown in the figure,the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.