The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then

A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$

A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $0$ to $3d$