The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. 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 $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distances 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 $3$ and between $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{C_1}{C_2}$ is

The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.