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 becomes $1/8^{th}$ of the original value. The value of the dielectric constant will be

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:

$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

$A$ parallel plate capacitor has a capacitance $C$ and the distance between the plates is $d$. If the space between the plates is filled with a dielectric material of dielectric constant $K$ as shown in the figure,find the new capacitance of the capacitor.

$A$ parallel-plate capacitor of capacitance $40 \mu F$ is connected to a $100 V$ power supply. Now,the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of the dielectric material,the extra charge and the change in the electrostatic energy in the capacitor,respectively,are:

$A$ capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when the medium between the plates is air. The charge on the plates becomes $100\,\mu C$ when the space between the plates is filled with oil. The dielectric constant of the oil is: