$A$ parallel plate capacitor is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

The plates of a parallel plate capacitor are charged up to $100 \,V$. $A$ $2 \,mm$ thick insulator sheet is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the insulator is

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in the figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q'_1$ and $Q'_2$. Then:

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

The plates of a parallel plate capacitor are separated by a distance $d.$ Two slabs of different dielectric constants $K_1$ and $K_2$ with thicknesses $\frac{3}{8} d$ and $\frac{d}{2}$ respectively are inserted into the capacitor. Due to this,the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25 K_2,$ find the value of $K_1.$