$A$ parallel plate capacitor has charge $5 \times 10^{-6} \ C$. $A$ dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \ C$, then the dielectric constant of the slab is . . . . . . .

The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

Four metallic plates $A, B, C$ and $D$ of the same size with the same separation between them are arranged as shown in the figure. Dielectric slabs of dielectric constant $K = 2$ are placed between $B, C$ and $C, D$ respectively. Plates $B$ and $D$ are connected together. The effective capacitance between $A$ and $C$ is (Assume capacitance of each pair of plates without dielectric is $C$):

The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is:

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

$A$ parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is