$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.

$A$ 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.

$A$ 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 $x = 3d$:

$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is:

$A$ parallel plate capacitor with width $4\,cm$,length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. $A$ dielectric slab of dielectric constant $5$ having length $1\,cm$,width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of the parallel plate capacitor. The electrostatic energy of this system will be ......... $\epsilon_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)