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

$A$ parallel plate capacitor with plate separation $5 \text{ mm}$ is charged by a battery. On introducing a mica sheet of $2 \text{ mm}$ thickness and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is . . . . . . .

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is: