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

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 $K_1.$

Four metallic plates $A, B, C$ and $D$ of the same size with the same separation between them are arranged as shown in the figure. Dielectric slabs of dielectric constant $K = 2$ are placed between $B, C$ and $C, D$ respectively. Plates $B$ and $D$ are connected together. The effective capacitance between $A$ and $C$ is (Assume capacitance of each pair of plates without dielectric is $C$):

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

$A$ slab of material with a dielectric constant of $3$ has the same area as the plates of a parallel plate capacitor but has a thickness of $\left(\frac{3}{4}\right) d$,where $d$ is the separation between the plates. What is the electrical potential difference between the plates when the slab is inserted? The initial electrical potential difference is $V_0$.

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is