$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.

In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. Then,

$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.

The function of a dielectric in a capacitor is

$A$ parallel plate capacitor has a plate area of $100\, m^{2}$ and a plate separation of $10\, m$. The space between the plates is filled up to a thickness of $5\, m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $'x'\, pF$. Given $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$,the value of $'x'$ to the nearest integer is:

$A$ parallel plate capacitor has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be: