In a parallel plate air capacitor of plate separation $d$,a dielectric slab of thickness $t$ is introduced between the plates $(t < d)$. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

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

The voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of $10^6\,V/m$. The plate area is $10^{-4}\,m^2$. What is the dielectric constant if the capacitance is $15\,pF$? (Given $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$)

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance:

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:

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