The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is $E_0$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections,the field between the plates shall be

$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)