A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,

When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates

A parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1,K_2,K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by 

The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ – 12}}farads$. The dielectric constant of wax is

Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distance as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k$ $=$ $2$. The capacitance of the system between $1$ and $3$ $\&$ $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{{{C_1}}}{{{C_2}}}$ is