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

$A$ 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 $x = 0$ to $x = 3d$:

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

An air capacitor has a capacitance of $1 \mu F$. Now the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is: [$d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the first and second dielectric respectively] (in $\mu F$)

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.