$A$ slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$,where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

$A$ parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor,which of the following does not change?

The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4 \text{ pF}$. The value of $x$ is

Two dielectric slabs having dielectric constants $K_1$ and $K_2$ of thicknesses $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates of a capacitor as shown in the figure. The net capacitance between $A$ and $B$ is [where $\varepsilon_0$ is the permittivity of free space].

$A$ parallel plate capacitor has a separation between plates of $0.885$ mm. It has a capacitance of $1$ $\mu$$F$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}$ $\Omega$m and resistance $17.7 \times 10^{14}$ $\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is . . . . . . . (Take permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ $F$/m)