Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

$A$ parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. $A$ dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

$A$ 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.

In a parallel plate capacitor,the separation between the plates is $3\,mm$ with air between them. Now,a $1\,mm$ thick layer of a material of dielectric constant $K = 2$ is introduced between the plates,due to which the capacity increases. In order to bring its capacity back to the original value,the separation between the plates must be increased to......$mm$. (in $.5$)

$A$ parallel plate capacitor with plate area $A$ and distance of separation $d$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$\varepsilon(x) = \varepsilon_{0} + kx$,for $(0 < x \leq d/2)$
$\varepsilon(x) = \varepsilon_{0} + k(d-x)$,for $(d/2 \leq x \leq d)$

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.