$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

$A$ parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$,as shown in the figures. The distance between the plates is $d$ and the area of each plate is $A$. If the capacitance in the first configuration and second configuration are $C_1$ and $C_2$ respectively,then $\frac{C_1}{C_2}$ is

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)

$A$ parallel plate capacitor is made of two plates of length $l$,width $w$ and separated by distance $d$. $A$ dielectric slab (dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when the dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$,then the force on the dielectric when it is near the edge is

Write the formula for the capacitance of a capacitor having a dielectric constant $K = 2$.